8x^2+62x=17

Solution for 8x^2+62x=17 equation:

8x^2+62x=17

We move all terms to the left:

8x^2+62x-(17)=0

a = 8; b = 62; c = -17;
Δ = b2-4ac
Δ = 622-4·8·(-17)
Δ = 4388
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4388}=\sqrt{4*1097}=\sqrt{4}*\sqrt{1097}=2\sqrt{1097}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-2\sqrt{1097}}{2*8}=\frac{-62-2\sqrt{1097}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+2\sqrt{1097}}{2*8}=\frac{-62+2\sqrt{1097}}{16}
$