Question

How to solve this I don’t know how to solve it myself

Answer 1

`-4x^2-40x=-156`

1. Write the equation in standard form:

`\begin{gathered} \text{Add 156 in both sides of the equation:} \\ -4x^2-40x+156=-156+156 \\ \\ -4x^2-40x+156=0 \end{gathered}`

2. Identify the coefficients a, b and c:

`\begin{gathered} ax^2+bx+c=0 \\ \\ \text{For the given equation:} \\ a=-4 \\ b=-40 \\ c=156 \end{gathered}`

3. Substitute the values into the quadratic equation:

`x=\frac{-(-40)\pm\sqrt[]{(-40)^2-4(-4)(156)}}{2(-4)}`

4. Solve the equation for x1 and x2:

`\begin{gathered} x_{}=\frac{-(-40)\pm\sqrt[]{(-40)^2-4(-4)(156)}}{2(-4)} \\ \\ x_{}=\frac{40\pm\sqrt[]{1600+2496}}{-8} \\ \\ x=\frac{40\pm\sqrt[]{4096}}{-8} \\ \\ x=(40\pm64)/(-8) \\ \\ \\ x_1=(40-64)/(-8)=(-24)/(-8)=3 \\ \\ x_2=(40+64)/(-8)=(104)/(-8)=-13 \end{gathered}`

Exact form and approximate form are the same for both solutions (x1 and x2):

`\begin{gathered} x_1=3 \\ \\ x_2=-13 \end{gathered}`