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x2+25x+10=20
We move all terms to the left:
x2+25x+10-(20)=0
We add all the numbers together, and all the variables
x^2+25x-10=0
a = 1; b = 25; c = -10;
Δ = b2-4ac
Δ = 252-4·1·(-10)
Δ = 665
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{665}}{2*1}=\frac{-25-\sqrt{665}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{665}}{2*1}=\frac{-25+\sqrt{665}}{2} $
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