The positive solution to the given quadratic equation is given by:
[tex]x=1.39[/tex]
The quadratic equation is given by:
[tex]2x^2+3x-8=0[/tex]
We know that the solution of the quadratic equation of the type:
[tex]ax^2+bx+c=0[/tex]
is given by:
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Here,
[tex]a=2,\ b=3\ and\ c=-8[/tex]
Hence, the solution is:
[tex]x=\dfrac{-3\pm \sqrt{3^2-4\times 2\times (-8)}}{2\times 2}\\\\x=\dfrac{-3\pm \sqrt{9+64}}{4}\\\\x=\dfrac{-3\pm \sqrt{73}}{4}\\\\x=\dfrac{-3+\sqrt{73}}{4}\ or\ x=\dfrac{-3-\sqrt{73}}{4}\\\\x=1.38600\ or\ x=-2.88600[/tex]
which on rounding to the nearest hundredth is:
[tex]x=1.39\ \ or\ \ x=-2.89[/tex]
Hence, the positive solution is:
[tex]x=1.39[/tex]
