Question

solve the quadratic equation
give your answer to 2 decimal places
: 3x^2+x-5=0

Answer 1

Given:

The quadratic equation is:

$3x^2+x-5=0$

To find:

The solution for the given equation rounded to 2 decimal places.

Solution:

Quadratic formula: If a quadratic equation is $ax^2+bx+c=0$, then:

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

We have,

$3x^2+x-5=0$

Here, $a=3,b=1,c=-5$. Using the quadratic formula, we get

$x=\dfrac{-1\pm \sqrt{1^2-4(3)(-5)}}{2(3)}$

$x=\dfrac{-1\pm \sqrt{1+60}}{6}$

$x=\dfrac{-1\pm \sqrt{61}}{6}$

$x=\dfrac{-1\pm 7.81025}{6}$

Now,

$x=\dfrac{-1+7.81025}{6}$

$x=1.13504167$

$x\approx 1.14$

And

$x=\dfrac{-1-7.81025}{6}$

$x=-1.468375$

$x\approx -1.47$

Therefore, the required solutions are 1.14 and -1.47.