5(x^2-1)=3(x-1)(x+2)

Solution for 5(x^2-1)=3(x-1)(x+2) equation:

5(x^2-1)=3(x-1)(x+2)

We move all terms to the left:

5(x^2-1)-(3(x-1)(x+2))=0

We multiply parentheses

5x^2-(3(x-1)(x+2))-5=0

We multiply parentheses ..

5x^2-(3(+x^2+2x-1x-2))-5=0


We calculate terms in parentheses: -(3(+x^2+2x-1x-2)), so:

3(+x^2+2x-1x-2)

We multiply parentheses

3x^2+6x-3x-6

We add all the numbers together, and all the variables

3x^2+3x-6

Back to the equation:

-(3x^2+3x-6)


We get rid of parentheses

5x^2-3x^2-3x+6-5=0

We add all the numbers together, and all the variables

2x^2-3x+1=0

a = 2; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·2·1
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*2}=\frac{2}{4}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*2}=\frac{4}{4}
=1
$