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15x^2-16x=240
We move all terms to the left:
15x^2-16x-(240)=0
a = 15; b = -16; c = -240;
Δ = b2-4ac
Δ = -162-4·15·(-240)
Δ = 14656
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{14656}=\sqrt{64*229}=\sqrt{64}*\sqrt{229}=8\sqrt{229}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{229}}{2*15}=\frac{16-8\sqrt{229}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{229}}{2*15}=\frac{16+8\sqrt{229}}{30} $
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