Question

1. solve for x. (3x² + 9)° (10x + 22)°

Explanation:

Step1: Recall property of parallelogram

In a parallelogram, adjacent - angles are supplementary, so $(3x^{2}+9)+(10x + 22)=180$.

Step2: Simplify the equation

Combine like - terms: $3x^{2}+10x+9 + 22=180$, which gives $3x^{2}+10x+31 = 180$. Then, subtract 180 from both sides to get the quadratic equation $3x^{2}+10x - 149 = 0$.

Step3: Use the quadratic formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 3$, $b = 10$, and $c=-149$. First, calculate the discriminant $\Delta=b^{2}-4ac=(10)^{2}-4\times3\times(-149)=100 + 1788 = 1888$. Then, $x=\frac{-10\pm\sqrt{1888}}{6}=\frac{-10\pm4\sqrt{118}}{6}=\frac{-5\pm2\sqrt{118}}{3}$.

Answer:

$x=\frac{-5 + 2\sqrt{118}}{3}$ or $x=\frac{-5 - 2\sqrt{118}}{3}$