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(12x^2+11x-4)=0
We get rid of parentheses
12x^2+11x-4=0
a = 12; b = 11; c = -4;
Δ = b2-4ac
Δ = 112-4·12·(-4)
Δ = 313
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{313}}{2*12}=\frac{-11-\sqrt{313}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{313}}{2*12}=\frac{-11+\sqrt{313}}{24} $
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