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2x^2+16x-96=0
a = 2; b = 16; c = -96;
Δ = b2-4ac
Δ = 162-4·2·(-96)
Δ = 1024
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{1024}=32$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-32}{2*2}=\frac{-48}{4} =-12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+32}{2*2}=\frac{16}{4} =4 $
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