[tex]-4x^2-40x=-156[/tex]
1. Write the equation in standard form:
[tex]\begin{gathered} \text{Add 156 in both sides of the equation:} \\ -4x^2-40x+156=-156+156 \\ \\ -4x^2-40x+156=0 \end{gathered}[/tex]
2. Identify the coefficients a, b and c:
[tex]\begin{gathered} ax^2+bx+c=0 \\ \\ \text{For the given equation:} \\ a=-4 \\ b=-40 \\ c=156 \end{gathered}[/tex]
3. Substitute the values into the quadratic equation:
[tex]x=\frac{-(-40)\pm\sqrt[]{(-40)^2-4(-4)(156)}}{2(-4)}[/tex]
4. Solve the equation for x1 and x2:
[tex]\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=\frac{40\pm64}{-8} \\ \\ \\ x_1=\frac{40-64}{-8}=\frac{-24}{-8}=3 \\ \\ x_2=\frac{40+64}{-8}=\frac{104}{-8}=-13 \end{gathered}[/tex]
Exact form and approximate form are the same for both solutions (x1 and x2):
[tex]\begin{gathered} x_1=3 \\ \\ x_2=-13 \end{gathered}[/tex]
