0=(2x^2-18x-812)

Solution for 0=(2x^2-18x-812) equation:

0=(2x^2-18x-812)

We move all terms to the left:

0-((2x^2-18x-812))=0

We add all the numbers together, and all the variables

-((2x^2-18x-812))=0


We calculate terms in parentheses: -((2x^2-18x-812)), so:

(2x^2-18x-812)

We get rid of parentheses

2x^2-18x-812

Back to the equation:

-(2x^2-18x-812)


We get rid of parentheses

-2x^2+18x+812=0

a = -2; b = 18; c = +812;
Δ = b2-4ac
Δ = 182-4·(-2)·812
Δ = 6820
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:

$\sqrt{\Delta}=\sqrt{6820}=\sqrt{4*1705}=\sqrt{4}*\sqrt{1705}=2\sqrt{1705}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{1705}}{2*-2}=\frac{-18-2\sqrt{1705}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{1705}}{2*-2}=\frac{-18+2\sqrt{1705}}{-4}
$