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5x^2-38x+16=0
a = 5; b = -38; c = +16;
Δ = b2-4ac
Δ = -382-4·5·16
Δ = 1124
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1124}=\sqrt{4*281}=\sqrt{4}*\sqrt{281}=2\sqrt{281}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{281}}{2*5}=\frac{38-2\sqrt{281}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{281}}{2*5}=\frac{38+2\sqrt{281}}{10} $
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