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3x^2+32x-19.4=0
a = 3; b = 32; c = -19.4;
Δ = b2-4ac
Δ = 322-4·3·(-19.4)
Δ = 1256.8
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{-(32)-\sqrt{1256.8}}{2*3}=\frac{-32-\sqrt{1256.8}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+\sqrt{1256.8}}{2*3}=\frac{-32+\sqrt{1256.8}}{6} $
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