If (m+8)(x^2)+m=7-8x has two real roots show that (m+9)(m-8)<0 where m is an arbitrary constant

For this we are going to test our knowledge of discriminats and factorisation. Firstly we will format the equation as (m+8)x²+8x+(m-7)=0From here we can see it takes the form ax²+bx+c, and as we know the equation has two real roots we know that the discriminant D is greater than 0. Therfore b²-4ac>0a=(m+8)    b=8     c=(m-7)8²-4(m+8)(m-7)>0   Pluggin in a,b,c64-4(m²+m-54)>0   expanding brackets and squaring 816-(m²+m-54)>0    dividing both sides by a factor of 40>(m²+m-54)-16     moving left handside to right handside by addition / subtraction0>m²+m-72        collecting terms0>(m+9)(m-8)      factorisingWe have now shown that (m+9)(m-8)<0 for the above equation when it has two real roots.