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25x^2-24x-48=0
a = 25; b = -24; c = -48;
Δ = b2-4ac
Δ = -242-4·25·(-48)
Δ = 5376
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5376}=\sqrt{256*21}=\sqrt{256}*\sqrt{21}=16\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-16\sqrt{21}}{2*25}=\frac{24-16\sqrt{21}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+16\sqrt{21}}{2*25}=\frac{24+16\sqrt{21}}{50} $
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