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malfutka [58]
3 years ago
13

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

Mathematics
2 answers:
Maru [420]3 years ago
8 0

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}

==========================================

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}    <em>simplify by 2</em>

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

Arisa [49]3 years ago
5 0

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

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