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5x^2-4x-19=0
a = 5; b = -4; c = -19;
Δ = b2-4ac
Δ = -42-4·5·(-19)
Δ = 396
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{11}}{2*5}=\frac{4-6\sqrt{11}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{11}}{2*5}=\frac{4+6\sqrt{11}}{10} $
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