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3x^2-110x+400=0
a = 3; b = -110; c = +400;
Δ = b2-4ac
Δ = -1102-4·3·400
Δ = 7300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{7300}=\sqrt{100*73}=\sqrt{100}*\sqrt{73}=10\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-110)-10\sqrt{73}}{2*3}=\frac{110-10\sqrt{73}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-110)+10\sqrt{73}}{2*3}=\frac{110+10\sqrt{73}}{6} $
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