(4x^2-36)=(3x^2-5x)

Solution for (4x^2-36)=(3x^2-5x) equation:

(4x^2-36)=(3x^2-5x)

We move all terms to the left:

(4x^2-36)-((3x^2-5x))=0

We get rid of parentheses

4x^2-((3x^2-5x))-36=0


We calculate terms in parentheses: -((3x^2-5x)), so:

(3x^2-5x)

We get rid of parentheses

3x^2-5x

Back to the equation:

-(3x^2-5x)


We get rid of parentheses

4x^2-3x^2+5x-36=0

We add all the numbers together, and all the variables

x^2+5x-36=0

a = 1; b = 5; c = -36;
Δ = b2-4ac
Δ = 52-4·1·(-36)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*1}=\frac{-18}{2}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*1}=\frac{8}{2}
=4
$