How do you solve $8x^2 + 4x = -5$ using the quadratic formula?

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Answer 1 · 2017-04-25T14:47:40

$x=(-1+-3i)/(4)$

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

where $ax^2+bx+c=0$

Here is your quadratic equation

$8x^2+4x=-5$

Move the $-5$ to the other side of the equation to complete your quadratic equation

$8x^2+4x+5=0$

Looking at $ax^2+bx+c=0$
I can see that your values are...

$a=8$
$b=4$
$c=5$

Now just put those values into the quadratic formula

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

$x=(-(4)+-sqrt((4)^2-4(8)(5)))/(2(8))$

$x=(-4+-sqrt(16-160))/(16)$

$x=(-4+-sqrt(-144))/(16)$

Here we notice that we have a negative square root.
This turns into an imaginary number or $i$ to remove the negative.

$x=(-4+-isqrt(144))/(16)$

$x=(-4+-12i)/(16)$

Notice we have a common factor of $4$ in the numerator and denominator.

$x=(4(-1+-3i))/(4(4))$

Cancel out common factor

$x=(cancel(4)(-1+-3i))/(cancel(4)(4))$

$x=(-1+-3i)/(4)$

How do you solve 8x^2 + 4x = -58x^2 + 4x = -5 using the quadratic formula?