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3(6x^2+5x-4)=0
We multiply parentheses
18x^2+15x-12=0
a = 18; b = 15; c = -12;
Δ = b2-4ac
Δ = 152-4·18·(-12)
Δ = 1089
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1089}=33$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-33}{2*18}=\frac{-48}{36} =-1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+33}{2*18}=\frac{18}{36} =1/2 $
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