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2x^2-52x+50=0
a = 2; b = -52; c = +50;
Δ = b2-4ac
Δ = -522-4·2·50
Δ = 2304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-48}{2*2}=\frac{4}{4} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+48}{2*2}=\frac{100}{4} =25 $
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