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16x^2+3x-425=0
a = 16; b = 3; c = -425;
Δ = b2-4ac
Δ = 32-4·16·(-425)
Δ = 27209
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{27209}=\sqrt{169*161}=\sqrt{169}*\sqrt{161}=13\sqrt{161}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13\sqrt{161}}{2*16}=\frac{-3-13\sqrt{161}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13\sqrt{161}}{2*16}=\frac{-3+13\sqrt{161}}{32} $
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