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

Answer:

Step-by-step 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

Answer 2

Answer:

$\large\boxed{x=\dfrac{-1-\sqrt3}{2}\ or\ x=\dfrac{-1+\sqrt3}{2}}$

Step-by-step 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 $x=\dfrac{-b}{2a}$

If Δ >0, then an equation has two real solutions $x=\dfrac{-b\pm\sqrt{\Delta}}{2a}$

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We have the equation:

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

Substitute:

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

$x=\dfrac{-2\pm2\sqrt3}{(2)(2)}=\dfrac{-2\pm2\sqrt3}{4}$    simplify by 2

$x=\dfrac{-1\pm\sqrt3}{2}$