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12x^2+8x-48=0
a = 12; b = 8; c = -48;
Δ = b2-4ac
Δ = 82-4·12·(-48)
Δ = 2368
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2368}=\sqrt{64*37}=\sqrt{64}*\sqrt{37}=8\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{37}}{2*12}=\frac{-8-8\sqrt{37}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{37}}{2*12}=\frac{-8+8\sqrt{37}}{24} $
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