0,4 Find the roots of the following quadratic equations using the quadratic formula
1) 6a²x² - 7abx - 3b²=0, a is not equal to zero

pls answer

b = b2-4ac

= (-7ab)^2 -4 (6a^2)(-3b^2)

= 49a^2b^2 + 72a^2b^2

= 121a^2b^2

Now, x = (7ab +- root 121a^2b^2)÷2(6a²)

= (7ab +- 11ab ) ÷ 12a²

= ( 20ab)/12a² or 4ab/12a²

= 5/3ab or - 1/3ab .

So, these are its roots .

Step-by-step explanation:

mx+89b(-98)nb

step-by-step explanation:

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.