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x2+10x=299
We move all terms to the left:
x2+10x-(299)=0
We add all the numbers together, and all the variables
x^2+10x-299=0
a = 1; b = 10; c = -299;
Δ = b2-4ac
Δ = 102-4·1·(-299)
Δ = 1296
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{1296}=36$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-36}{2*1}=\frac{-46}{2} =-23 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+36}{2*1}=\frac{26}{2} =13 $
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