2x^2+16x-58=O

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

2x^2+16x-58=

We move all terms to the left:

2x^2+16x-58-()=0

We add all the numbers together, and all the variables

2x^2+16x=0

a = 2; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·2·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*2}=\frac{-32}{4}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*2}=\frac{0}{4}
=0
$