Solve 4y^2-8y+6

$$4 { y }^{ 2 } -8y+6$$

Factor

$$2\left(2y^{2}-4y+3\right)$$

Solution Steps

Factor out 2. Polynomial 2y^{2}-4y+3 is not factored since it does not have any rational roots.

$$2\left(2y^{2}-4y+3\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$4y^{2}-8y+6=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 4\times 6}}{2\times 4}$$

Square -8.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\times 4\times 6}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-\left(-8\right)±\sqrt{64-16\times 6}}{2\times 4}$$

Multiply -16 times 6.

$$y=\frac{-\left(-8\right)±\sqrt{64-96}}{2\times 4}$$

Add 64 to -96.

$$y=\frac{-\left(-8\right)±\sqrt{-32}}{2\times 4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$4y^{2}-8y+6$$

Evaluate

$$4y^{2}-8y+6$$