5(3x^2-x-4)=x(3x-5)+88

Solution for 5(3x^2-x-4)=x(3x-5)+88 equation:

5(3x^2-x-4)=x(3x-5)+88

We move all terms to the left:

5(3x^2-x-4)-(x(3x-5)+88)=0

We multiply parentheses

15x^2-5x-(x(3x-5)+88)-20=0


We calculate terms in parentheses: -(x(3x-5)+88), so:

x(3x-5)+88

We multiply parentheses

3x^2-5x+88

Back to the equation:

-(3x^2-5x+88)


We get rid of parentheses

15x^2-3x^2-5x+5x-88-20=0

We add all the numbers together, and all the variables

12x^2-108=0

a = 12; b = 0; c = -108;
Δ = b2-4ac
Δ = 02-4·12·(-108)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-72}{2*12}=\frac{-72}{24}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+72}{2*12}=\frac{72}{24}
=3
$