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2x^2+16x-13=5
We move all terms to the left:
2x^2+16x-13-(5)=0
We add all the numbers together, and all the variables
2x^2+16x-18=0
a = 2; b = 16; c = -18;
Δ = b2-4ac
Δ = 162-4·2·(-18)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-20}{2*2}=\frac{-36}{4} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+20}{2*2}=\frac{4}{4} =1 $
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