Question

How do you solve $y=2{{x}^{2}}+5x-1$ using the quadratic formula?

Answer 1

From the question given, we have been asked to solve $y=2{{x}^{2}}+5x-1$ using the quadratic formula. We can solve the above given quadratic equation in the question by using the quadratic formula by substituting the values of coefficients of the quadratic equation in the quadratic formula and then we have to simplify it further to get the final roots for the given quadratic equation in the given question.

Complete answer:
From the question, it has been already given that $y=2{{x}^{2}}+5x-1$
We can clearly observe that the given quadratic equation is in the form of the general form of the quadratic equation that is $a{{x}^{2}}+bx+c=0$.
By comparing the coefficients of the given quadratic equation in the given question and the general form of the quadratic equation, we get the values of coefficients as,

& a=2 \\\ & b=5 \\\ & c=-1 \\\ \end{aligned}$$ Now, we have got the values of coefficients by comparing the coefficients. Now, what we have to do is substitute the values of coefficients in the quadratic formula. Quadratic formula is shown below: $$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ By substituting the values of coefficients in the above quadratic formula, we get $$x=\dfrac{-5\pm \sqrt{{{5}^{2}}-4\left( 2 \right)\left( -1 \right)}}{2\left( 2 \right)}$$ On furthermore simplifying, we get $$x=\dfrac{-5\pm \sqrt{25-\left( -8 \right)}}{4}$$ $$\Rightarrow x=\dfrac{-5\pm \sqrt{25+8}}{4}$$ $$\Rightarrow x=\dfrac{-5\pm \sqrt{33}}{4}$$ Therefore, we got the roots for the given quadratic equation by using the quadratic formula. **Note:** We should be very careful while writing the quadratic formula because if you write the formula wrong then the whole question will go wrong. Also, we should be very careful while doing the calculation especially when solving the equation by using the quadratic formula. Also, we should be very careful while substituting the values of coefficients. Similarly we can solve any expression as follows $$y=4{{x}^{2}}+4x+1\Rightarrow y={{\left( 2x+1 \right)}^{2}}\Rightarrow x=\pm \left( \dfrac{-1}{2} \right)$$