11x^2+3x+18=6x^2-18(4x-1)

Solution for 11x^2+3x+18=6x^2-18(4x-1) equation:

11x^2+3x+18=6x^2-18(4x-1)

We move all terms to the left:

11x^2+3x+18-(6x^2-18(4x-1))=0


We calculate terms in parentheses: -(6x^2-18(4x-1)), so:

6x^2-18(4x-1)

We multiply parentheses

6x^2-72x+18

Back to the equation:

-(6x^2-72x+18)


We get rid of parentheses

11x^2-6x^2+3x+72x-18+18=0

We add all the numbers together, and all the variables

5x^2+75x=0

a = 5; b = 75; c = 0;
Δ = b2-4ac
Δ = 752-4·5·0
Δ = 5625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5625}=75$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(75)-75}{2*5}=\frac{-150}{10}
=-15
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(75)+75}{2*5}=\frac{0}{10}
=0
$