Question

Find the real solutions of the quadratic equation $10y^2-16y+5=0$.

Short answer

The given equation has two distinct real solutions $x=\frac{8+\sqrt{14}}{10},\frac{8-\sqrt{14}}{10}$.

Step-by-step solution

Step 1. By method of quadratic formula.

The roots of the quadratic equation of the form $a{x}^2+bx+c=0$where $a\ne 0$can be evaluated by using the quadratic formula shown below.

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

Step 2. Find roots of quadratic equation by quadratic formula.

The given quadratic equation is $10y^2-16y+5=0$whose roots can be evaluated by quadratic formula as shown below.

$$\begin{gathered} x=\frac{-\left(-16\right)\pm \sqrt{{\left(-16\right)}^2-4\left(10\right)\left(5\right)}}{2\left(10\right)} \\ =\frac{16\pm \sqrt{56}}{20} \\ =\frac{16\pm 2\sqrt{14}}{20} \\ =\frac{8+\sqrt{14}}{10},\frac{8-\sqrt{14}}{10} \end{gathered}$$

Step 3. Description of steps.

The roots of the given quadratic equation are real, i.e. $x=\frac{8+\sqrt{14}}{10},\frac{8-\sqrt{14}}{10}$.