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9x^2-4x-5=0
a = 9; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·9·(-5)
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14}{2*9}=\frac{-10}{18} =-5/9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14}{2*9}=\frac{18}{18} =1 $
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