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2x^2+4x-261=0
a = 2; b = 4; c = -261;
Δ = b2-4ac
Δ = 42-4·2·(-261)
Δ = 2104
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2104}=\sqrt{4*526}=\sqrt{4}*\sqrt{526}=2\sqrt{526}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{526}}{2*2}=\frac{-4-2\sqrt{526}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{526}}{2*2}=\frac{-4+2\sqrt{526}}{4} $
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