how do i solve the equation -2x^2 - 3x + 5 = 0 with quadratic formula?

Question

1. -5 and 1

2. -2 and 1

3. -5/2 and 1

4. 5/2 and -1

Answer 1

its easier just to factorise -(2x^2+3x-5)=(2x-5)(x+1)
so 4. 5/2 and -1

Answer 2

It might be easier to just factor this, but the problem specifically asks you to use the quadratic forumula.

The quadratic formula says that if you have a quadratic equation in the form of ax² + bx + c = 0, then the solution is
x = [ -b ±√(b² - 4ac) ] / 2a

So here, a = -2, b= -3, and c = 5. Plug these into the above equation, and calculate x:
x = [ 3 ±√(9 - 4(-2)(5)) ] / 2(-2)
x = [ 3 ±√(9 + 40) ] / -4
x = [ 3 ±√(49) ] / -4
x = [ 3 ± 7 ] / -4
x = [ 3+7] / -4, [3-7]/-4
x = -5/2, 1

Answer 3

Any equation with the form ax^2 + bx + c = 0 can be solved using the equation

x = (-b +/- sqrt(b^2 - 4ac))/2a

a = -2
b = -3
c = 5

x = (3 +/- sqrt((-3)^2 - 4 * -2 * 5))/(2 * -2)
x = (3 +/- sqrt(9 + 40))/-4
x = (3 +/- sqrt(49))/-4
x = (3 +/- 7)/-4
x = 10/-4 or -4/-4
x = -5/2 or 1

Answer 4

a = -2, b = -3, c = 5, so
2a = -4, -b = 3, b² = 9, -4ac = -4(-2)(5) = 40

x = 3/-4 ± √(9 + 40) / -4
x = -3/4 ± 7/4
x = -3/4 + 7/4 = 4/4 = 1 or
x = -3/4 - 7/4 = -10/4 = -5/2

Answer 5

-2x^2 - 3x + 5 = 0 multiply by -1
2x^2 + 3x -5=0
(2x+5)(x-1)=0
x= -5/2
or
x= 1

3. -5/2 and 1