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21x^2+25x=8
We move all terms to the left:
21x^2+25x-(8)=0
a = 21; b = 25; c = -8;
Δ = b2-4ac
Δ = 252-4·21·(-8)
Δ = 1297
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{1297}}{2*21}=\frac{-25-\sqrt{1297}}{42} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{1297}}{2*21}=\frac{-25+\sqrt{1297}}{42} $
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