Question

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Answer 1

Answer:

$\boxed{\textsf{ The roots of the quadratic equation is \textbf{ (-4) and 3 }.}}$

Step-by-step explanation:

A quadratic equation is given to us . Here we aren't given what to do with it. So , lets find out the roots of the equation . The given quadratic equation is :-

$\sf\implies 4x^2+x-3 =0$

Now with respect to Standard form ax²+bx + c . The roots of the equation can be find out by the quadratic formula also known as " Shreedharyacharya's Formula " .

Quadratic Formula :-

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

Now with respect to Standard Form we have ,

• a = 4 •b = 1 •c = (-3)

Put on the respective values :-

$\sf\implies x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\\\\sf\implies x =\dfrac{-1\pm\sqrt{1^2-4(1)(-3)}}{2(1)}\\\\\sf\implies x =\dfrac{-1\pm \sqrt{1+48}}{2}\\\\\sf\implies x =\dfrac{-1\pm \sqrt{49}}{2}\\\\\sf\implies x = \dfrac{-1\pm 7 }{2}\\\\\sf\implies x =\dfrac{-1-7}{2},\dfrac{-1+7}{2}\\\\\sf\implies x =\dfrac{-8}{2},\dfrac{6}{2}\\\\\sf\implies \boxed{\pink{\sf x = (-4) , 3 }}$