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X^2+50X+48=0
a = 1; b = 50; c = +48;
Δ = b2-4ac
Δ = 502-4·1·48
Δ = 2308
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2308}=\sqrt{4*577}=\sqrt{4}*\sqrt{577}=2\sqrt{577}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{577}}{2*1}=\frac{-50-2\sqrt{577}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{577}}{2*1}=\frac{-50+2\sqrt{577}}{2} $
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