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(B) The given quadratic equation is $x^{2}-3x-4=0$.Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=-3, c=-4$.The quadratic formul

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$-1, 4$

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

Solve the following equation by using the quadratic formula,if the equation has a solution in $R$: $x^{2}-3x-4=0$.

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