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4x^2+19=-20x
We move all terms to the left:
4x^2+19-(-20x)=0
We get rid of parentheses
4x^2+20x+19=0
a = 4; b = 20; c = +19;
Δ = b2-4ac
Δ = 202-4·4·19
Δ = 96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{6}}{2*4}=\frac{-20-4\sqrt{6}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{6}}{2*4}=\frac{-20+4\sqrt{6}}{8} $
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