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-12+5x^2-4x=0
a = 5; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·5·(-12)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-16}{2*5}=\frac{-12}{10} =-1+1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+16}{2*5}=\frac{20}{10} =2 $
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