1,080=(2x^2+10x-64)

Solution for 1,080=(2x^2+10x-64) equation:

1.080=(2x^2+10x-64)

We move all terms to the left:

1.080-((2x^2+10x-64))=0

We add all the numbers together, and all the variables

-((2x^2+10x-64))+1.08=0


We calculate terms in parentheses: -((2x^2+10x-64)), so:

(2x^2+10x-64)

We get rid of parentheses

2x^2+10x-64

Back to the equation:

-(2x^2+10x-64)


We get rid of parentheses

-2x^2-10x+64+1.08=0

We add all the numbers together, and all the variables

-2x^2-10x+65.08=0

a = -2; b = -10; c = +65.08;
Δ = b2-4ac
Δ = -102-4·(-2)·65.08
Δ = 620.64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-\sqrt{620.64}}{2*-2}=\frac{10-\sqrt{620.64}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{620.64}}{2*-2}=\frac{10+\sqrt{620.64}}{-4}
$