(4x^2-24)=(3x^2-5x)

Solution for (4x^2-24)=(3x^2-5x) equation:

(4x^2-24)=(3x^2-5x)

We move all terms to the left:

(4x^2-24)-((3x^2-5x))=0

We get rid of parentheses

4x^2-((3x^2-5x))-24=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-24=0

We add all the numbers together, and all the variables

x^2+5x-24=0

a = 1; b = 5; c = -24;
Δ = b2-4ac
Δ = 52-4·1·(-24)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*1}=\frac{6}{2}
=3
$