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6x+2x^2=95
We move all terms to the left:
6x+2x^2-(95)=0
a = 2; b = 6; c = -95;
Δ = b2-4ac
Δ = 62-4·2·(-95)
Δ = 796
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{796}=\sqrt{4*199}=\sqrt{4}*\sqrt{199}=2\sqrt{199}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{199}}{2*2}=\frac{-6-2\sqrt{199}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{199}}{2*2}=\frac{-6+2\sqrt{199}}{4} $
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