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2x(3x-5)=9
We move all terms to the left:
2x(3x-5)-(9)=0
We multiply parentheses
6x^2-10x-9=0
a = 6; b = -10; c = -9;
Δ = b2-4ac
Δ = -102-4·6·(-9)
Δ = 316
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{79}}{2*6}=\frac{10-2\sqrt{79}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{79}}{2*6}=\frac{10+2\sqrt{79}}{12} $
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