(3x^2-50)=(2x^2-5x)

Solution for (3x^2-50)=(2x^2-5x) equation:

(3x^2-50)=(2x^2-5x)

We move all terms to the left:

(3x^2-50)-((2x^2-5x))=0

We get rid of parentheses

3x^2-((2x^2-5x))-50=0


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

(2x^2-5x)

We get rid of parentheses

2x^2-5x

Back to the equation:

-(2x^2-5x)


We get rid of parentheses

3x^2-2x^2+5x-50=0

We add all the numbers together, and all the variables

x^2+5x-50=0

a = 1; b = 5; c = -50;
Δ = b2-4ac
Δ = 52-4·1·(-50)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-15}{2*1}=\frac{-20}{2}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+15}{2*1}=\frac{10}{2}
=5
$