(9x^2+8x)=(2x^2+3x)

Solution for (9x^2+8x)=(2x^2+3x) equation:

(9x^2+8x)=(2x^2+3x)

We move all terms to the left:

(9x^2+8x)-((2x^2+3x))=0

We get rid of parentheses

9x^2+8x-((2x^2+3x))=0


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

(2x^2+3x)

We get rid of parentheses

2x^2+3x

Back to the equation:

-(2x^2+3x)


We get rid of parentheses

9x^2-2x^2+8x-3x=0

We add all the numbers together, and all the variables

7x^2+5x=0

a = 7; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·7·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*7}=\frac{-10}{14}
=-5/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*7}=\frac{0}{14}
=0
$