Question

how to solve the above quadratic equation​

Answer 1

Question:

$\sf{R^2+3R-40=0}$

Solution:

$\sf{\bold{Solving\:it\:by\:discriminant\:method}}$

$\sf{For\:general\:quadratic\:equation\:ax^2+bx+c=0}$

$\sf{Discriminant=D=b^2-4ac}$

$\sf{Therefore\:Comparing\:it\:with\:the\:given\:equation,}\\\sf{we\:get,}$

$\sf{a=1,b=3,c=-40}$

$\sf{D=(3)^2-4(1)(-40)}\\\sf{D=9+160}\\\sf{D=169}$

$\sf{we\:know\:that,}$

$\sf{R=\dfrac{-b\pm\sqrt{D}}{2a}}$

$\sf{R=\dfrac{-5\pm\sqrt{169}}{2}}$

$\sf{R=\dfrac{-5\pm13}{2}}$

$\sf{By\:solving,\:we\:get\colon}$

$\boxed{\sf{R=4\:\:or\:-9}}$