2x^2+7x-2=0?

Answer 1

Hello Dear Joshua ☺;

2x ² + 7x - 2 = 0

if a = 2 , b = 7 & c = -2
∆ = b ² - 4ac
∆ = ( 7 ² ) - 4 ( 2)(-2)= 49 + 16
∆ = 65 & ∆ ≥ 0
x 1 = ( -b + √ ∆ ) / 2a = (-(7) + √ 65) / 4 = -7 + √ 65 / 4
x 2 = ( -b - √ ∆ ) / 2a = (-(7) - √ 65) / 4 = -7 - √ 65 / 4
x1 ≈ -7 + 8.06 / 4 = 15.06 / 4 = 3.765
x2 ≈ -7 - 8.06 / 4 = - 1.06 / 4 = -0.265

Good Luck Dear ♣ ♣ ♣

Answer 2

find 2 numbers whose product is -4 and sum is 7.

No integer numbers seem to does this, so we will resort to
completing the square:

x^2 + 7/2 X - 1 = O

x^2 + 7/2 X + 49/16 - 49/16 -16/16 = 0

(x + 7/4)^2 = 65/16

x = -7/4 +/- sqrt(65/16)

please check for errors

Answer 3

2x^2 + 7x - 2 = 0

Assume you want to solve this equation.

It doesn't factor easily, so you will have to apply the quadratic formula

x = [- b +- sqrt(b^2 - 4ac)]/2a

Where a = 2, b = 7, c = -2

Answer 4

2x^2 + 7x - 2 = 0
x^2 + 7/2x - 1 = 0
x^2 + 7/2x +(7/4)^2 - (7/4)^2 - 1=0
x^2 + 7/2x +(7/4)^2=1+ 49/16 = 65/16
(x + 7/4)^2 = 65/16
x + 7/4 = +/- sqrt(65)/4
x = (-7 + sqrt(65))/4
x = (-7 - sqrt(65))/4

Answer 5

Use the quadratic equation: [-b(+or-)sqrt(b^2-4ac)]/2a for equations of the form ax^2+bx+c=0. Which of course your problem is.
so what do we get:
[-7(+or-)sqrt((7^2-(4)(2)(-2))]/2(2)
which equals [-7(+or-)sqrt(65)]/4.

Answer 6

using the Quadratic formula
x=[-7+/-rt(49+16)]/4
=[-7+/-rt(65)]/2

Answer 7

Use quadratic formula:

x=[ -7 + or - root (7^2-4*2*(-2)]/(2*2)

x = [-7 + or - root (49 +16)]/4

x = (-7 +65^.5)/4
x = (-7 -65^.5)/4

Answer 8

for ax^2+bx+c the roots are
x =[ -b +-sq rt(b^2-4ac)]/2a
here a=2 b=7 c=-2
x=[ -7 +-sq rt(7^2+16)]/4
= [-7 +-sq rt(65)]/4
= [-7 +-8.06)]/4
so x = -15.06/4 or +1.06/4
= -3.765 or 0.265

Answer 9

x = (-7 +/- sqrt(65))/4

by the quadratic formula

Answer 10

x=(-7 +/- sqrt(65))/4