18x^2+27x-5=(3x+5)

Solution for 18x^2+27x-5=(3x+5) equation:

18x^2+27x-5=(3x+5)

We move all terms to the left:

18x^2+27x-5-((3x+5))=0


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

(3x+5)

We get rid of parentheses

3x+5

Back to the equation:

-(3x+5)


We get rid of parentheses

18x^2+27x-3x-5-5=0

We add all the numbers together, and all the variables

18x^2+24x-10=0

a = 18; b = 24; c = -10;
Δ = b2-4ac
Δ = 242-4·18·(-10)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-36}{2*18}=\frac{-60}{36}
=-1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+36}{2*18}=\frac{12}{36}
=1/3
$