Question

find the roots of 2x^2-3x-5 using quardatic formula​

Answer 1

Step-by-step explanation:

2-3-5=0

-bt - 4ac

The quadratic formula is z =

2a

This equation is in standard quadratic form, so

we know that:

a = 2.b = - 3, andc =- 5.

Plug them into the equation and solve for æ:

-(-3) + y-3)- 4(2)(- 5)

2(2)

3 + V9+40

3+ 49

10

and

4

5

5and-1

Hope this helps!

v refers to root*

Answer 2

Answer:

the answer is (2.5 , -1 )

Step-by-step explanation:

2x^2-3x-5        { a=2 ; b=-3 ; c=-5 }

quadrant formula = -b ± √ b² -4ac ÷ 2a

substitute the values in the formula

-(-3) ±√(-3)²-4(2)(-5)÷2(2)

3±√9-(-40)÷4

3±√9+40÷4

3±√49÷4

3±√49÷4

3±7÷4                    ( ∴ √49 = 7 )

3+7÷4 ; 3-7÷4

10÷4 ; -4÷4

then ( 2.5 , -1 )              (or)   u can leave  10÷4 directly