Question

Solve the equation $5x^2-3x-4=0$ and give your answer to correct two decimals.

Answer 1

Step 1: Compare the equation with the general equation

The given equation is $5x^2-3x-4=0$

comparing the equation with the general equation $a{x}^2+bx+c=0$, we get

$a=5,b=-3,c=-4$

Step 2: Find the discriminant of the equation

$$\begin{gathered} d=b^2-4ac \\ d={(-3)}^2-4\times \left(5\right)\times (-4) \\ d=9+80 \\ d=89 \end{gathered}$$

Step 3: Find the roots of the equation

$$\begin{gathered} x=\frac{-b\pm \sqrt{d}}{2a} \\ x=\frac{-(-3)\pm \sqrt{89}}{2\times 5} \\ x=\frac{3\pm 9.4339}{10} \\ x=\frac{12.4339}{10}\mathrm{or}x=\frac{-6.4339}{10} \\ x=1.24339\mathrm{or}x=-0.64339 \\ x\approx 1.243\mathrm{or}x\approx -0.643 \end{gathered}$$

Hence the solution of the equation $5x^2-3x-4=0$ is $x\approx 1.243\mathrm{or}x\approx -0.643$.