5(x^2+16x+80)=4(x^2+22x+121)

Solution for 5(x^2+16x+80)=4(x^2+22x+121) equation:

5(x^2+16x+80)=4(x^2+22x+121)

We move all terms to the left:

5(x^2+16x+80)-(4(x^2+22x+121))=0

We multiply parentheses

5x^2+80x-(4(x^2+22x+121))+400=0


We calculate terms in parentheses: -(4(x^2+22x+121)), so:

4(x^2+22x+121)

We multiply parentheses

4x^2+88x+484

Back to the equation:

-(4x^2+88x+484)


We get rid of parentheses

5x^2-4x^2+80x-88x-484+400=0

We add all the numbers together, and all the variables

x^2-8x-84=0

a = 1; b = -8; c = -84;
Δ = b2-4ac
Δ = -82-4·1·(-84)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-20}{2*1}=\frac{-12}{2}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+20}{2*1}=\frac{28}{2}
=14
$