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x^2=-26x
We move all terms to the left:
x^2-(-26x)=0
We get rid of parentheses
x^2+26x=0
a = 1; b = 26; c = 0;
Δ = b2-4ac
Δ = 262-4·1·0
Δ = 676
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{676}=26$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-26}{2*1}=\frac{-52}{2} =-26 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+26}{2*1}=\frac{0}{2} =0 $
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