Question

(15 points ) PLEASE HELP ASAP! Also please show your work! Solve.

Use the quadratic formula. Write solutions in simplist radical form. 2x^2+2x-1=0

Answer 1

`\large\boxed{x=(-1-\sqrt3)/(2)\ or\ x=(-1+\sqrt3)/(2)}`

Explanation:

The quadratic formula of a quadratic equation:

`ax^2+bx+c=0`

Discriminant of a Quadratic is
`\Delta=b^2-4ac`

If Δ < 0, then an equation has no real solution (has two complex solutions)

If Δ = 0, then an equation has one real solution
`x=(-b)/(2a)`

If Δ >0, then an equation has two real solutions
`x=(-b\pm√(\Delta))/(2a)`

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We have the equation:

`2x^2+2x-1=0\\\\a=2,\ b=2,\ c=-1`

Substitute:

`\Delta=2^2-4(2)(-1)=4+8=12>0\\\\\sqrt\Delta=√(12)=√(4\cdot3)=\sqrt4\cdot\sqrt3=2\sqrt3`

`x=(-2\pm2\sqrt3)/((2)(2))=(-2\pm2\sqrt3)/(4)` simplify by 2

`x=(-1\pm\sqrt3)/(2)`

Answer 2

Explanation:

ax^2+bx+c

-b +- sqaure root of b^2 -4ac/2a

-2 +- square root (2)^2-4(2)(-1)/2(2)

-2 +- square root 4+8 /4

-2 +- 2 square root 3 /4

reduce

-1 +- square root 3/2