I need help solving for x in quadratic equation... x^2=-7x+4?

Answer 1

x=(-7+sqrt[65])/2

or

x=(-7-sqrt[65])/2

Answer 2

Completing the square

x² = - 7x + 4

x² + 7x = - 7x + 4 + 7x

x² + 7x = 4

x² + 7x +_____= 4 +____

x² + 7x + 49/4 = 4 + 49/4

(x + 7/2)(x + 7/2) = 16/4 + 49/4

(x + 7/2)² = 65/4

(√x + 7/2)² = ± √65 / √4

x + 7/2 = ± √65 / 2

x + 7/2 - 7/2 = - 7/2 ± √65 / 2

x = - 7/2 ± 8.062257748 / 2

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Solving for +

x = - 7/2 + 8.062257748 / 2

x = 1.062257748 / 2

x = 0.531128874

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Solving for -

x = - 7/2 - 8.062257748 / 2

x = - 15.06225775 / 2

x = - 7.531128874

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The quadratic formula also works

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Answer 3

First, make it look like a quadratatic {ax^2+bx+c=0}

x^2+7x-4=0

Then apply the quadratic formula to find the roots...

(-7+sqrt(49+16))/2 and (-7-sqrt(49+16))/2

approx 0.531 and -7.531

Answer 4

x^2+7x-4=0 (general form of a quadratic equation)
then, by using a calculator, x=0.53 or x=-7.53

Sorry, there is no whole number for this equation.

Answer 5

x^2 + 7x - 4 = 0
x^2 + 7x + (7/2)^2 - (7/2)^2 - 4 =0
(x + 7/2)^2 - (49 + 16)/4 = 0
(x + 7/2 -sqrt65/2)(x + 7/2 + sqrt65/2) = 0

x = -7/2 +/- sqrt65/2

Answer 6

you will need the general formula to solve.

x = (b +- sqrt(b^2 - 4ac))/2a, where a, b and c are coefficient of ax^2 + bx + c = 0; so a = 1, b = 7 and c = -4

Plug the coefficients a,b and c into the equation, you will obtain x = 0.53 or -7.53

Answer 7

x^2=-7x+4
x^2 + 7x - 4 = 0

x = (-b +- sqrt( b^2 - 4ac))/2a
a = 1
b = 7
c = -4

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

That's it!