Application of quadratic equation of one variable (urgent)! In 2006, a real estate company invested 60 million yuan in real estate business. At the beginning of 2007, the company took the sum of the capital invested in 2006 and profits as the total capital. As a result, the interest rate in 2007 was 10% higher than that in 2006, and the total capital in 2007 was 93.6 million yuan. How many million yuan was the actual profit in 2007

Application of quadratic equation of one variable (urgent)! In 2006, a real estate company invested 60 million yuan in real estate business. At the beginning of 2007, the company took the sum of the capital invested in 2006 and profits as the total capital. As a result, the interest rate in 2007 was 10% higher than that in 2006, and the total capital in 2007 was 93.6 million yuan. How many million yuan was the actual profit in 2007

Suppose that the yield rate in 2006 is x (6000 + 60000x) + (6000 + 60000x) (x + 10%) = 9360
After calculating x, the actual profit in 2007 is (x + 10%) (6000 + 6000x)

On the application of quadratic equation of one variable
1: It is known that the length of a rectangle is twice the side length of a square, the width is 2 cm more than the side length of the square, and the area is 96 square cm larger than the area of the square. Find the side length of the square and the length and width of the rectangle
2: In the middle of a 55 meter long and 45 meter wide rectangular green space, build two mutually perpendicular paths of the same width. The remaining area that can be used for greening is 2000 square meters. Calculate the width of the path

Let the side length of the square be x, then the length of the rectangle is 2x, the width is x + 2, and the area of the rectangle is 2x (x + 2);
The square of square area X; 2x (x + 2) - x * x = 96; the solution is x = 8, so the length of rectangle is 16 and the width is 10
Let the road width be X
55*45-55*2x=2000,x=4.5

Use one variable quadratic equation!
It is known that the lengths of AB and AC on both sides of △ ABC are two real roots of the univariate quadratic equation x & # 178; - (2k + 3) x + K & # 178; + 3K + 2 = 0, and the length of the third side BC is 5. When k is the value, AB & # 178; + AC & # 178; = BC & # 178;?

According to the meaning of the title
[x-（k+1）][x-（k+2）]=0,
The solution is X1 = K + 1, X2 = K + 2,
If △ ABC is a right triangle and BC is a hypotenuse,
Then there is (K + 1) 2 + (K + 2) 2 = 52,
The solution is K1 = 2, K2 = - 5,
∴k=2
I hope my answer is useful to you, and I hope it can be accepted as a satisfactory answer in time

Application problems related to quadratic equation of one variable
Hongda car rental company has 120 taxis, and the daily rent of each taxi is 160 yuan. The taxi business is in short supply every day. In order to meet the market demand, with the approval of Youguang department, the company plans to increase the daily rent appropriately. According to the market survey, for every 10 yuan increase in the daily rent of a car, the number of taxis per day will be reduced by 6 correspondingly, When the company increases the daily rental of each car by 10 yuan, the total daily rental income of the company will be 19380 yuan?
Let X be increased

If the rent is increased by X Yuan, the daily rent income of the company can reach 19380 yuan
（160+X）[120-（X/10）*6]=19380
The results are as follows
X^2-40X+300=0
X1=10
X2=30
A: when the rent is increased by 10 yuan or 30 yuan, the daily rent income can reach 19380 yuan
Remember a title in the book?

Who can help me solve the quadratic equation of one variable? Quick
（35-2x)x=150

（35-2x)x=150
2x²-35x+150=0
(2x-15)(x-10)=0
x1=15/2,x2=10