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7x+x^2=400
We move all terms to the left:
7x+x^2-(400)=0
a = 1; b = 7; c = -400;
Δ = b2-4ac
Δ = 72-4·1·(-400)
Δ = 1649
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{-(7)-\sqrt{1649}}{2*1}=\frac{-7-\sqrt{1649}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{1649}}{2*1}=\frac{-7+\sqrt{1649}}{2} $
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