2x^2+16x-58=0

Solution for 2x^2+16x-58=0 equation:

2x^2+16x-58=0

a = 2; b = 16; c = -58;
Δ = b2-4ac
Δ = 162-4·2·(-58)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-12\sqrt{5}}{2*2}=\frac{-16-12\sqrt{5}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+12\sqrt{5}}{2*2}=\frac{-16+12\sqrt{5}}{4}
$