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(2x^2+10x-28)=0
We get rid of parentheses
2x^2+10x-28=0
a = 2; b = 10; c = -28;
Δ = b2-4ac
Δ = 102-4·2·(-28)
Δ = 324
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{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-18}{2*2}=\frac{-28}{4} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+18}{2*2}=\frac{8}{4} =2 $
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