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4x^2+81x+400=0
a = 4; b = 81; c = +400;
Δ = b2-4ac
Δ = 812-4·4·400
Δ = 161
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(81)-\sqrt{161}}{2*4}=\frac{-81-\sqrt{161}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(81)+\sqrt{161}}{2*4}=\frac{-81+\sqrt{161}}{8} $
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