Answer:
Step-by-step explanation:
2x^2 + 3x =9
when equation like a * (x^2)+ b*x + c =0
solution like;
x= (-b +sqrt(b^2 - 4ac))/2a
and
x= (-b - sqrt(b^2 - 4ac))/2a
Now consider 2x^2 + 3x -9 =0
so a=2, b=3, c= (-9)
Let first consider x= (-b + sqrt(b^2 - 4ac))/2a;
x= (-b + sqrt(b^2 - 4ac))/2a
x= ( -3 + sqrt( 3^2 - 4*2*(-9)) )/2 *2
= 6/4
= 1.5
Let second consider x= (-b - sqrt(b^2 - 4ac))/2a;
x= (-b - sqrt(b^2 - 4ac))/2a
x= ( -3 - sqrt( 3^2 - 4*2*(-9)) )/2 *2
= -12/4
= -3
