Question

Use the quadratic formula to find both solutions to the quadratic equation given below. 2x^2+3x-5=0

Answer 1

Answer:

$x =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

Where a = 2 , b= 3, c= -5, replacing we have this:

$x =\frac{-3 \pm \sqrt{(-3)^2 -4(2)(-5)}}{2*2}$

And simplifying we got:

$x = \frac{-3 \pm \sqrt{49}}{4}$

And the two solutions are:

$x_1 = \frac{-3+7}{4}= 1$

$x_2 = \frac{-3-7}{4}= -\frac{5}{2}$

And the correct options are:

B and C

Step-by-step explanation:

We have the following equation given:

$2x^2 +3x -5=0$

And if we use the quadratic formula given by:

$x =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

Where a = 2 , b= 3, c= -5, replacing we have this:

$x =\frac{-3 \pm \sqrt{(-3)^2 -4(2)(-5)}}{2*2}$

And simplifying we got:

$x = \frac{-3 \pm \sqrt{49}}{4}$

And the two solutions are:

$x_1 = \frac{-3+7}{4}= 1$

$x_2 = \frac{-3-7}{4}= -\frac{5}{2}$

And the correct options are:

B and C