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16x^2+25x+9=0
a = 16; b = 25; c = +9;
Δ = b2-4ac
Δ = 252-4·16·9
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-7}{2*16}=\frac{-32}{32} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+7}{2*16}=\frac{-18}{32} =-9/16 $
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