2x(x-7)+36=4(2x-5)+60

Solution for 2x(x-7)+36=4(2x-5)+60 equation:

2x(x-7)+36=4(2x-5)+60

We move all terms to the left:

2x(x-7)+36-(4(2x-5)+60)=0

We multiply parentheses

2x^2-14x-(4(2x-5)+60)+36=0


We calculate terms in parentheses: -(4(2x-5)+60), so:

4(2x-5)+60

We multiply parentheses

8x-20+60

We add all the numbers together, and all the variables

8x+40

Back to the equation:

-(8x+40)


We get rid of parentheses

2x^2-14x-8x-40+36=0

We add all the numbers together, and all the variables

2x^2-22x-4=0

a = 2; b = -22; c = -4;
Δ = b2-4ac
Δ = -222-4·2·(-4)
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{129}}{2*2}=\frac{22-2\sqrt{129}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{129}}{2*2}=\frac{22+2\sqrt{129}}{4}
$