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2x^2+4x-1296=0
a = 2; b = 4; c = -1296;
Δ = b2-4ac
Δ = 42-4·2·(-1296)
Δ = 10384
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{10384}=\sqrt{16*649}=\sqrt{16}*\sqrt{649}=4\sqrt{649}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{649}}{2*2}=\frac{-4-4\sqrt{649}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{649}}{2*2}=\frac{-4+4\sqrt{649}}{4} $
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