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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


We calculate terms in parentheses: -((9x+18)3), so:

(9x+18)3

We multiply parentheses

27x+54

Back to the equation:

-(27x+54)


We add all the numbers together, and all the variables

3x^2-(27x+54)+9=0

We get rid of parentheses

3x^2-27x-54+9=0

We add all the numbers together, and all the variables

3x^2-27x-45=0

a = 3; b = -27; c = -45;
Δ = b2-4ac
Δ = -272-4·3·(-45)
Δ = 1269
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:

$\sqrt{\Delta}=\sqrt{1269}=\sqrt{9*141}=\sqrt{9}*\sqrt{141}=3\sqrt{141}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-3\sqrt{141}}{2*3}=\frac{27-3\sqrt{141}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+3\sqrt{141}}{2*3}=\frac{27+3\sqrt{141}}{6}
$