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2x^2+40x-192=0
a = 2; b = 40; c = -192;
Δ = b2-4ac
Δ = 402-4·2·(-192)
Δ = 3136
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{3136}=56$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-56}{2*2}=\frac{-96}{4} =-24 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+56}{2*2}=\frac{16}{4} =4 $
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