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15x^2-8=4x
We move all terms to the left:
15x^2-8-(4x)=0
a = 15; b = -4; c = -8;
Δ = b2-4ac
Δ = -42-4·15·(-8)
Δ = 496
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{31}}{2*15}=\frac{4-4\sqrt{31}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{31}}{2*15}=\frac{4+4\sqrt{31}}{30} $
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