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4x^2+24x-26=0
a = 4; b = 24; c = -26;
Δ = b2-4ac
Δ = 242-4·4·(-26)
Δ = 992
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{992}=\sqrt{16*62}=\sqrt{16}*\sqrt{62}=4\sqrt{62}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{62}}{2*4}=\frac{-24-4\sqrt{62}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{62}}{2*4}=\frac{-24+4\sqrt{62}}{8} $
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