Answer
$x=8$ and $x=-1$
Work Step by Step
$2x^{2}-14x+10=26$
Subtracting $10$ from both sides we get
$2x^{2}-14x=16$
Divide both sides of the equation by $2$ to obtain
$x^{2}-7x=8$
Add $(\frac{b}{2})^{2}=(\frac{-7}{2})^{2}=(\frac{7}{2})^{2}$ to both sides to complete the square.
$x^{2}-7x+(\frac{7}{2})^{2}=8+(\frac{7}{2})^{2}$
$\implies (x-\frac{7}{2})^{2}=8+\frac{49}{4}$
$\implies (x-\frac{7}{2})^{2}=\frac{81}{4}$
Taking square root on both sides, we have
$x-\frac{7}{2}=\pm\frac{9}{2}$
$\implies x=\frac{7}{2}\pm \frac{9}{2}$
The solutions of the given equation are $x=\frac{7}{2}+\frac{9}{2}=8$ and $x=\frac{7}{2}-\frac{9}{2}=-1$
