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2x^2+12x-224=0
a = 2; b = 12; c = -224;
Δ = b2-4ac
Δ = 122-4·2·(-224)
Δ = 1936
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{1936}=44$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-44}{2*2}=\frac{-56}{4} =-14 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+44}{2*2}=\frac{32}{4} =8 $
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