how do you solve 4x^2+8x+7=4

Answer 1

You need to first subtract 4 from both sides of the equation and see if you can factor the quadratic equation:

4x² + 8x + 3 = 0

x = (-8 ± √(64 - (4)(4)(3))/(4•2) = -1 ± √(64 - 48)/8 = -1 ± (1/8)(4) = -3/2, -1/2

Checking

(x +3/2)(x+1/2) = x² + x/2 + 3x/2 + 3/4 = x² + 2x + 3/4 = 0 --> 4x² + 8x + 3 = 0 √

Adding 4 to each side gives the original equation

Answer 2

Hi Alexis

If you want to solve for x

4x^2+8x+7=4

You can move everything over to one side of the equation

4x² + 8x + 7 - 4 =0

Leaving

4x² + 8x + 3 = 0

This can be factored

(2x + 3)(2x + 1) =0

x = -3/2 and x = -1/2

These solutions are the x intercepts. You can plug them back into your original question to check

4x^2+8x+7=4

4(-3/2)² + 8(-3/2) + 7 = 4

4(9/4) - 12 + 7 =4

9 - 12 + 7 = 4

-3 + 7 = 4

4 = 4

And

4(-1/2)² + 8(-1/2) + 7 = 4

4(1/4) - 4 + 7 = 4

1 - 4 + 7 = 4

-3 + 7 = 4

4 = 4

You can graph your parabola at Desmos.com as well