The positive solution to the quadratic equation [tex]2 x^{2}+3 x-8=0[/tex] is x = 1.39
Given quadratic equation is [tex]2 x^{2}+3 x-8=0[/tex]
The general quadratic equation is of form:
[tex]a x^{2}+b x+c=0[/tex]
Now comparing the general equation with the given equation we get
a = 2 , b = 3 and c = -8
The formula to determine roots of the quadratic equation is:
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
On plugging in vlaues, we get
[tex]x=\frac{-3 \pm \sqrt{3^{2}-4 \times 2 \times(-8)}}{2 \times 2}[/tex]
[tex]\mathrm{x}=\frac{-3 \pm \sqrt{9-(-64)}}{4}[/tex]
On solving we get,
[tex]\begin{array}{l}{\mathrm{x}=\frac{-3 \pm \sqrt{9+64}}{4}} \\\\ {\mathrm{x}=\frac{-3 \pm \sqrt{73}}{4}}\end{array}[/tex]
[tex]\mathrm{x}=\frac{-3+\sqrt{73}}{4} \text { OR } \mathrm{x}=\frac{-3-\sqrt{73}}{4}[/tex]
x = 1.39 OR x = -2.89
Hence , the positive solution to the quadratic equation is x = 1.39