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9x^2+10x=18
We move all terms to the left:
9x^2+10x-(18)=0
a = 9; b = 10; c = -18;
Δ = b2-4ac
Δ = 102-4·9·(-18)
Δ = 748
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{748}=\sqrt{4*187}=\sqrt{4}*\sqrt{187}=2\sqrt{187}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{187}}{2*9}=\frac{-10-2\sqrt{187}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{187}}{2*9}=\frac{-10+2\sqrt{187}}{18} $
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