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6x(3x+5)=36
We move all terms to the left:
6x(3x+5)-(36)=0
We multiply parentheses
18x^2+30x-36=0
a = 18; b = 30; c = -36;
Δ = b2-4ac
Δ = 302-4·18·(-36)
Δ = 3492
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3492}=\sqrt{36*97}=\sqrt{36}*\sqrt{97}=6\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-6\sqrt{97}}{2*18}=\frac{-30-6\sqrt{97}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+6\sqrt{97}}{2*18}=\frac{-30+6\sqrt{97}}{36} $
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