3(x2+5)-6=(9x+18)

Solution for 3(x2+5)-6=(9x+18) equation:

3(x2+5)-6=(9x+18)

We move all terms to the left:

3(x2+5)-6-((9x+18))=0

We add all the numbers together, and all the variables

3(+x^2+5)-((9x+18))-6=0

We multiply parentheses

3x^2-((9x+18))+15-6=0


We calculate terms in parentheses: -((9x+18)), so:

(9x+18)

We get rid of parentheses

9x+18

Back to the equation:

-(9x+18)


We add all the numbers together, and all the variables

3x^2-(9x+18)+9=0

We get rid of parentheses

3x^2-9x-18+9=0

We add all the numbers together, and all the variables

3x^2-9x-9=0

a = 3; b = -9; c = -9;
Δ = b2-4ac
Δ = -92-4·3·(-9)
Δ = 189
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{189}=\sqrt{9*21}=\sqrt{9}*\sqrt{21}=3\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{21}}{2*3}=\frac{9-3\sqrt{21}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{21}}{2*3}=\frac{9+3\sqrt{21}}{6}
$