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45x^2-4x-1=0
a = 45; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·45·(-1)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14}{2*45}=\frac{-10}{90} =-1/9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14}{2*45}=\frac{18}{90} =1/5 $
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