Answer:
[tex] x =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
Where a = 2 , b= 3, c= -5, replacing we have this:
[tex]x =\frac{-3 \pm \sqrt{(-3)^2 -4(2)(-5)}}{2*2}[/tex]
And simplifying we got:
[tex] x = \frac{-3 \pm \sqrt{49}}{4}[/tex]
And the two solutions are:
[tex] x_1 = \frac{-3+7}{4}= 1[/tex]
[tex] x_2 = \frac{-3-7}{4}= -\frac{5}{2}[/tex]
And the correct options are:
B and C
Step-by-step explanation:
We have the following equation given:
[tex] 2x^2 +3x -5=0[/tex]
And if we use the quadratic formula given by:
[tex] x =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
Where a = 2 , b= 3, c= -5, replacing we have this:
[tex]x =\frac{-3 \pm \sqrt{(-3)^2 -4(2)(-5)}}{2*2}[/tex]
And simplifying we got:
[tex] x = \frac{-3 \pm \sqrt{49}}{4}[/tex]
And the two solutions are:
[tex] x_1 = \frac{-3+7}{4}= 1[/tex]
[tex] x_2 = \frac{-3-7}{4}= -\frac{5}{2}[/tex]
And the correct options are:
B and C
