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3x(5x-4)=56
We move all terms to the left:
3x(5x-4)-(56)=0
We multiply parentheses
15x^2-12x-56=0
a = 15; b = -12; c = -56;
Δ = b2-4ac
Δ = -122-4·15·(-56)
Δ = 3504
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3504}=\sqrt{16*219}=\sqrt{16}*\sqrt{219}=4\sqrt{219}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{219}}{2*15}=\frac{12-4\sqrt{219}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{219}}{2*15}=\frac{12+4\sqrt{219}}{30} $
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