30=2(x-3)(x-5)

Solution for 30=2(x-3)(x-5) equation:

30=2(x-3)(x-5)

We move all terms to the left:

30-(2(x-3)(x-5))=0

We multiply parentheses ..

-(2(+x^2-5x-3x+15))+30=0


We calculate terms in parentheses: -(2(+x^2-5x-3x+15)), so:

2(+x^2-5x-3x+15)

We multiply parentheses

2x^2-10x-6x+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+30=0

We add all the numbers together, and all the variables

-2x^2+16x=0

a = -2; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-2)·0
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{-(16)-16}{2*-2}=\frac{-32}{-4}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-2}=\frac{0}{-4}
=0
$