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12x^2+28x-80=0
a = 12; b = 28; c = -80;
Δ = b2-4ac
Δ = 282-4·12·(-80)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-68}{2*12}=\frac{-96}{24} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+68}{2*12}=\frac{40}{24} =1+2/3 $
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