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3x(3x+2)=50
We move all terms to the left:
3x(3x+2)-(50)=0
We multiply parentheses
9x^2+6x-50=0
a = 9; b = 6; c = -50;
Δ = b2-4ac
Δ = 62-4·9·(-50)
Δ = 1836
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1836}=\sqrt{36*51}=\sqrt{36}*\sqrt{51}=6\sqrt{51}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{51}}{2*9}=\frac{-6-6\sqrt{51}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{51}}{2*9}=\frac{-6+6\sqrt{51}}{18} $
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