-2+4a-8=(a+5)(a-2)

Solution for -2+4a-8=(a+5)(a-2) equation:

-2+4a-8=(a+5)(a-2)

We move all terms to the left:

-2+4a-8-((a+5)(a-2))=0

We add all the numbers together, and all the variables

4a-((a+5)(a-2))-10=0

We multiply parentheses ..

-((+a^2-2a+5a-10))+4a-10=0


We calculate terms in parentheses: -((+a^2-2a+5a-10)), so:

(+a^2-2a+5a-10)

We get rid of parentheses

a^2-2a+5a-10

We add all the numbers together, and all the variables

a^2+3a-10

Back to the equation:

-(a^2+3a-10)


We add all the numbers together, and all the variables

4a-(a^2+3a-10)-10=0

We get rid of parentheses

-a^2+4a-3a+10-10=0

We add all the numbers together, and all the variables

-1a^2+a=0

a = -1; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-1)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-1}=\frac{-2}{-2}
=1
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-1}=\frac{0}{-2}
=0
$