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5x^2-4x=+9
We move all terms to the left:
5x^2-4x-(+9)=0
We add all the numbers together, and all the variables
5x^2-4x-9=0
a = 5; b = -4; c = -9;
Δ = b2-4ac
Δ = -42-4·5·(-9)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14}{2*5}=\frac{-10}{10} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14}{2*5}=\frac{18}{10} =1+4/5 $
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