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12x^2+36x+26=0
a = 12; b = 36; c = +26;
Δ = b2-4ac
Δ = 362-4·12·26
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{3}}{2*12}=\frac{-36-4\sqrt{3}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{3}}{2*12}=\frac{-36+4\sqrt{3}}{24} $
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