63=(2x+10)(x+3)

Solution for 63=(2x+10)(x+3) equation:

63=(2x+10)(x+3)

We move all terms to the left:

63-((2x+10)(x+3))=0

We multiply parentheses ..

-((+2x^2+6x+10x+30))+63=0


We calculate terms in parentheses: -((+2x^2+6x+10x+30)), so:

(+2x^2+6x+10x+30)

We get rid of parentheses

2x^2+6x+10x+30

We add all the numbers together, and all the variables

2x^2+16x+30

Back to the equation:

-(2x^2+16x+30)


We get rid of parentheses

-2x^2-16x-30+63=0

We add all the numbers together, and all the variables

-2x^2-16x+33=0

a = -2; b = -16; c = +33;
Δ = b2-4ac
Δ = -162-4·(-2)·33
Δ = 520
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:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{130}}{2*-2}=\frac{16-2\sqrt{130}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{130}}{2*-2}=\frac{16+2\sqrt{130}}{-4}
$