(s-6)(s-3)+(2s-7)(s+12)=s(2s-1)-44

Solution for (s-6)(s-3)+(2s-7)(s+12)=s(2s-1)-44 equation:

(s-6)(s-3)+(2s-7)(s+12)=s(2s-1)-44

We move all terms to the left:

(s-6)(s-3)+(2s-7)(s+12)-(s(2s-1)-44)=0

We multiply parentheses ..

(+s^2-3s-6s+18)+(2s-7)(s+12)-(s(2s-1)-44)=0


We calculate terms in parentheses: -(s(2s-1)-44), so:

s(2s-1)-44

We multiply parentheses

2s^2-1s-44

Back to the equation:

-(2s^2-1s-44)


We get rid of parentheses

s^2-2s^2-3s-6s+(2s-7)(s+12)+1s+18+44=0

We multiply parentheses ..

s^2-2s^2+(+2s^2+24s-7s-84)-3s-6s+1s+18+44=0

We add all the numbers together, and all the variables

-1s^2+(+2s^2+24s-7s-84)-8s+62=0

We get rid of parentheses

-1s^2+2s^2+24s-7s-8s-84+62=0

We add all the numbers together, and all the variables

s^2+9s-22=0

a = 1; b = 9; c = -22;
Δ = b2-4ac
Δ = 92-4·1·(-22)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-13}{2*1}=\frac{-22}{2}
=-11
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+13}{2*1}=\frac{4}{2}
=2
$