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40=9x+10x^2
We move all terms to the left:
40-(9x+10x^2)=0
We get rid of parentheses
-10x^2-9x+40=0
a = -10; b = -9; c = +40;
Δ = b2-4ac
Δ = -92-4·(-10)·40
Δ = 1681
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{1681}=41$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-41}{2*-10}=\frac{-32}{-20} =1+3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+41}{2*-10}=\frac{50}{-20} =-2+1/2 $
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