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2x^2+25x-8=5
We move all terms to the left:
2x^2+25x-8-(5)=0
We add all the numbers together, and all the variables
2x^2+25x-13=0
a = 2; b = 25; c = -13;
Δ = b2-4ac
Δ = 252-4·2·(-13)
Δ = 729
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{729}=27$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-27}{2*2}=\frac{-52}{4} =-13 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+27}{2*2}=\frac{2}{4} =1/2 $
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