(5x^2+3x)=(3x+20)

Solution for (5x^2+3x)=(3x+20) equation:

(5x^2+3x)=(3x+20)

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -((3x+20)), so:

(3x+20)

We get rid of parentheses

3x+20

Back to the equation:

-(3x+20)


We get rid of parentheses

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

We add all the numbers together, and all the variables

5x^2-20=0

a = 5; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·5·(-20)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*5}=\frac{-20}{10}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*5}=\frac{20}{10}
=2
$