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15x^2-38x+5=0
a = 15; b = -38; c = +5;
Δ = b2-4ac
Δ = -382-4·15·5
Δ = 1144
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1144}=\sqrt{4*286}=\sqrt{4}*\sqrt{286}=2\sqrt{286}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{286}}{2*15}=\frac{38-2\sqrt{286}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{286}}{2*15}=\frac{38+2\sqrt{286}}{30} $
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