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45x^2+28x+4=0
a = 45; b = 28; c = +4;
Δ = b2-4ac
Δ = 282-4·45·4
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-8}{2*45}=\frac{-36}{90} =-2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+8}{2*45}=\frac{-20}{90} =-2/9 $
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