# Thread: solving 4x^2 - 12x + 25 = 0

1. ## solving 4x^2 - 12x + 25 = 0

can someone help me solve 4x^2 - 12x + 25 = 0? please

2. $
x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}
$

a= 4
b= -12
c= 25

3. Originally Posted by osoares10
can someone help me solve this? please
$\Delta = b^2-4ac = (-12)^2-(4)(4)(25) = -256=-\left(16^2\right)$

This means your solutions will be complex and it would be best to slap it straight into the quadratic formula:

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

• $a = 4$
• $b = -12$
• $c = 25$

4. The equation for any quadratic equation follows the form $ax^2+bx+c$; what we need to do here is find the factors of a and c, that, when multiplied, will give us b.

The first thing we want to do when solving a quadratic, is to determine whether or not there is a real solution to the problem. We do this by taking a look at the quadratic formula:

$x=\frac{-b-\sqrt{b^2-4ac}}{2a}$

And noting that there is only a real solution if the expression under the square root, is greater than 0. We see though that:

$(-12)^2-4(4)(25) \Longrightarrow 144-400 < 0$

Thus the equation has no real solution. If, from where you are in your instruction, you are familiar with complex roots, you must go further and solve the equation using the quadratic formula. If not, it will suffice to simply say because the discriminant is less than zero, there is no real solution to your equation.