2x^2+16x=17

Solution for 2x^2+16x=17 equation:

2x^2+16x=17

We move all terms to the left:

2x^2+16x-(17)=0

a = 2; b = 16; c = -17;
Δ = b2-4ac
Δ = 162-4·2·(-17)
Δ = 392
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:

$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-14\sqrt{2}}{2*2}=\frac{-16-14\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14\sqrt{2}}{2*2}=\frac{-16+14\sqrt{2}}{4}
$