(a-6)(a-12)+(2a-5)(a-9)=a(2a-35)+157

Solution for (a-6)(a-12)+(2a-5)(a-9)=a(2a-35)+157 equation:

(a-6)(a-12)+(2a-5)(a-9)=a(2a-35)+157

We move all terms to the left:

(a-6)(a-12)+(2a-5)(a-9)-(a(2a-35)+157)=0

We multiply parentheses ..

(+a^2-12a-6a+72)+(2a-5)(a-9)-(a(2a-35)+157)=0


We calculate terms in parentheses: -(a(2a-35)+157), so:

a(2a-35)+157

We multiply parentheses

2a^2-35a+157

Back to the equation:

-(2a^2-35a+157)


We get rid of parentheses

a^2-2a^2-12a-6a+(2a-5)(a-9)+35a+72-157=0

We multiply parentheses ..

a^2-2a^2+(+2a^2-18a-5a+45)-12a-6a+35a+72-157=0

We add all the numbers together, and all the variables

-1a^2+(+2a^2-18a-5a+45)+17a-85=0

We get rid of parentheses

-1a^2+2a^2-18a-5a+17a+45-85=0

We add all the numbers together, and all the variables

a^2-6a-40=0

a = 1; b = -6; c = -40;
Δ = b2-4ac
Δ = -62-4·1·(-40)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-14}{2*1}=\frac{-8}{2}
=-4
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+14}{2*1}=\frac{20}{2}
=10
$