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12x^2+11x-5=0
a = 12; b = 11; c = -5;
Δ = b2-4ac
Δ = 112-4·12·(-5)
Δ = 361
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*12}=\frac{-30}{24} =-1+1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*12}=\frac{8}{24} =1/3 $
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