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3x^2+60x+23=0
a = 3; b = 60; c = +23;
Δ = b2-4ac
Δ = 602-4·3·23
Δ = 3324
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3324}=\sqrt{4*831}=\sqrt{4}*\sqrt{831}=2\sqrt{831}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-2\sqrt{831}}{2*3}=\frac{-60-2\sqrt{831}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+2\sqrt{831}}{2*3}=\frac{-60+2\sqrt{831}}{6} $
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