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6x^2+16x-35=0
a = 6; b = 16; c = -35;
Δ = b2-4ac
Δ = 162-4·6·(-35)
Δ = 1096
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1096}=\sqrt{4*274}=\sqrt{4}*\sqrt{274}=2\sqrt{274}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{274}}{2*6}=\frac{-16-2\sqrt{274}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{274}}{2*6}=\frac{-16+2\sqrt{274}}{12} $
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