50=x2+((2x-5)*2)

Solution for 50=x2+((2x-5)*2) equation:

50=x2+((2x-5)*2)

We move all terms to the left:

50-(x2+((2x-5)*2))=0


We calculate terms in parentheses: -(x2+((2x-5)*2)), so:

x2+((2x-5)*2)

We add all the numbers together, and all the variables

x^2+((2x-5)*2)


We calculate terms in parentheses: +((2x-5)*2), so:

(2x-5)*2

We multiply parentheses

4x-10

Back to the equation:

+(4x-10)


We get rid of parentheses

x^2+4x-10

Back to the equation:

-(x^2+4x-10)


We get rid of parentheses

-x^2-4x+10+50=0

We add all the numbers together, and all the variables

-1x^2-4x+60=0

a = -1; b = -4; c = +60;
Δ = b2-4ac
Δ = -42-4·(-1)·60
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{-(-4)-16}{2*-1}=\frac{-12}{-2}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+16}{2*-1}=\frac{20}{-2}
=-10
$