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4x^2+36x+65=0
a = 4; b = 36; c = +65;
Δ = b2-4ac
Δ = 362-4·4·65
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-16}{2*4}=\frac{-52}{8} =-6+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+16}{2*4}=\frac{-20}{8} =-2+1/2 $
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