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10x^2-48x-576=0
a = 10; b = -48; c = -576;
Δ = b2-4ac
Δ = -482-4·10·(-576)
Δ = 25344
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{25344}=\sqrt{2304*11}=\sqrt{2304}*\sqrt{11}=48\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-48\sqrt{11}}{2*10}=\frac{48-48\sqrt{11}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48\sqrt{11}}{2*10}=\frac{48+48\sqrt{11}}{20} $
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