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3x^2+44x+24=0
a = 3; b = 44; c = +24;
Δ = b2-4ac
Δ = 442-4·3·24
Δ = 1648
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1648}=\sqrt{16*103}=\sqrt{16}*\sqrt{103}=4\sqrt{103}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-4\sqrt{103}}{2*3}=\frac{-44-4\sqrt{103}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+4\sqrt{103}}{2*3}=\frac{-44+4\sqrt{103}}{6} $
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