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3x^2-46x+90=0
a = 3; b = -46; c = +90;
Δ = b2-4ac
Δ = -462-4·3·90
Δ = 1036
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1036}=\sqrt{4*259}=\sqrt{4}*\sqrt{259}=2\sqrt{259}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-46)-2\sqrt{259}}{2*3}=\frac{46-2\sqrt{259}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-46)+2\sqrt{259}}{2*3}=\frac{46+2\sqrt{259}}{6} $
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