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3x^2+6x-448=0
a = 3; b = 6; c = -448;
Δ = b2-4ac
Δ = 62-4·3·(-448)
Δ = 5412
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5412}=\sqrt{4*1353}=\sqrt{4}*\sqrt{1353}=2\sqrt{1353}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{1353}}{2*3}=\frac{-6-2\sqrt{1353}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{1353}}{2*3}=\frac{-6+2\sqrt{1353}}{6} $
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