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2x^2-20x+49=19
We move all terms to the left:
2x^2-20x+49-(19)=0
We add all the numbers together, and all the variables
2x^2-20x+30=0
a = 2; b = -20; c = +30;
Δ = b2-4ac
Δ = -202-4·2·30
Δ = 160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{10}}{2*2}=\frac{20-4\sqrt{10}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{10}}{2*2}=\frac{20+4\sqrt{10}}{4} $
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