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4x^2-96x+288=0
a = 4; b = -96; c = +288;
Δ = b2-4ac
Δ = -962-4·4·288
Δ = 4608
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4608}=\sqrt{2304*2}=\sqrt{2304}*\sqrt{2}=48\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-48\sqrt{2}}{2*4}=\frac{96-48\sqrt{2}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+48\sqrt{2}}{2*4}=\frac{96+48\sqrt{2}}{8} $
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