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x2-20x+52=0

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Solution for x2-20x+52=0 equation: 



x2-20x+52=0

We add all the numbers together, and all the variables

x^2-20x+52=0

a = 1; b = -20; c = +52;
Δ = b2-4ac
Δ = -202-4·1·52
Δ = 192
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{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}


x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{3}}{2*1}=\frac{20-8\sqrt{3}}{2}



x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{3}}{2*1}=\frac{20+8\sqrt{3}}{2}

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