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-x^2+20x+28=0
We add all the numbers together, and all the variables
-1x^2+20x+28=0
a = -1; b = 20; c = +28;
Δ = b2-4ac
Δ = 202-4·(-1)·28
Δ = 512
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16\sqrt{2}}{2*-1}=\frac{-20-16\sqrt{2}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16\sqrt{2}}{2*-1}=\frac{-20+16\sqrt{2}}{-2} $
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