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(D) Given equation is $-2x^{2} + 3x + 2 = 0$.On comparing with the standard form $ax^{2} + bx + c = 0$,we get:$a = -2, b = 3, c = 2$.The discriminant

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$-\frac{1}{2}, 2$

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(D) Given equation is $-2x^{2} + 3x + 2 = 0$.On comparing with the standard form $ax^{2} + bx + c = 0$,we get:$a = -2, b = 3, c = 2$.The discriminant $D$ is given by $D = b^{2} - 4ac$.$D = (3)^{2} - 4(-2)(2) = 9 + 16 = 25$.Since $D > 0$,the equation has two distinct real roots.Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:$x = \frac{-3 \pm \sqrt{25}}{2(-2)} = \frac{-3 \pm 5}{-4}$.Case $1$: $x = \frac{-3 + 5}{-4} = \frac{2}{-4} = -\frac{1}{2}$.Case $2$: $x = \frac{-3 - 5}{-4} = \frac{-8}{-4} = 2$.Thus,the roots are $-\frac{1}{2}$ and $2$.

Find whether the following equation has real roots. If real roots exist,find them.
$-2x^{2} + 3x + 2 = 0$

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Find whether the following equation has real roots. If real roots exist,find them.$-2x^{2} + 3x + 2 = 0$