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6x^2+56x-18=0
a = 6; b = 56; c = -18;
Δ = b2-4ac
Δ = 562-4·6·(-18)
Δ = 3568
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3568}=\sqrt{16*223}=\sqrt{16}*\sqrt{223}=4\sqrt{223}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-4\sqrt{223}}{2*6}=\frac{-56-4\sqrt{223}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+4\sqrt{223}}{2*6}=\frac{-56+4\sqrt{223}}{12} $
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