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12x^2+40x-63=0
a = 12; b = 40; c = -63;
Δ = b2-4ac
Δ = 402-4·12·(-63)
Δ = 4624
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{4624}=68$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-68}{2*12}=\frac{-108}{24} =-4+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+68}{2*12}=\frac{28}{24} =1+1/6 $
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