Quadratic formula?

Question

Can you help me solve this. I've tried and I can't seem to get the answer.

2x^ + 4x - 7 = 0

I'm supposed to use the quadratic formula.

Answer 1

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

x = ( -4 +/- sqrt(4^2 - 4*2*-7) / 2*2 )

x = (-4 +/- sqrt( 16 + 56 ) / 4

x = (-4 +/- sqrt(72) ) / 4

Answer 2

Let's try the quadratic formula
ax² + bx + c = 0
x = (-b +- √(b² - 4ac))/2a
x = (-4 +- √(16 +56))/4
x = (-4 +- √(72))/4
x = (-4 +- 6√(2))/4
x = (-2 + 3√(2))/2, x = (-2 - 3√(2))/2
x = 1.12132, x = -3.1213
.

Answer 3

formula so -b+or - sqaureroot of [(b)squared-4ac]/2a
2x^ + 4x - 7 = 0
pick a=2 b=4 c=-7
[-4+or- squareroot of [ (4)squared -- 4*2*-7]/ 2*--7
[-4+or - squareroot of [ (16) +56]/ -14
[-4+-squareroot of 72 ]/ -14
and then you would need a calculator to get the square root of 72 add or it by -4 and divide by -14
and you get your answer

Answer 4

[-4 +- sqrroot(16-(2(2)(-7)) ] / [2(2)]

the answer probably wont turn out perfect...just plug in the values into the quadratic formula. A=2 B=4 C=(-7)

[-b +- sqrroot(b^2 - (4ac)) ] / 2a

Answer 5

x = [- 4 ± √(16 + 56)] / 4
x = [- 4 ±√(72)] / 4
x = [- 4 ± 6√2 ] / 4
x = - 1 ± (3/2).√2

Answer 6

-4 +/- sqrt(16+56) / 4

(-4+/- 6sqrt 2) /4

x= (-2+3sqrt2)/2 OR (-2-3sqrt2)/2