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4x^2+21+20x=0
a = 4; b = 20; c = +21;
Δ = b2-4ac
Δ = 202-4·4·21
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8}{2*4}=\frac{-28}{8} =-3+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8}{2*4}=\frac{-12}{8} =-1+1/2 $
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