Respuesta :
To solve for y, apply the distributive property:
[tex]10y + 12 = 24[/tex]
Then, subtract 12 from both sides and you will get:
[tex]10y = 12 \\ [/tex]
Then solve for y, by dividing both sides by 10.
Your final answer will be y=1.2
[tex]10y + 12 = 24[/tex]
Then, subtract 12 from both sides and you will get:
[tex]10y = 12 \\ [/tex]
Then solve for y, by dividing both sides by 10.
Your final answer will be y=1.2
The correct answers would be [tex] \frac{-6 +/- 2\sqrt{6} }{5} [/tex]
We can find this answer by first putting it into standard form. To do this we need to multiply the parenthesis and then solve for 0.
(5y + 6)^2 = 24
25y^2 + 60y + 36 = 24
25y^2 + 60y + 12 = 0
Now we can find the a, b and c values for the quadratic equation based on the standard form.
a = 25 (number attached to x^2)
b = 60 (number attached to x)
c = 12 (number with no variable)
Now we use these in the quadratic equation and simplify.
[tex] \frac{-b +/- \sqrt{ b^{2} -4ac } }{2a} [/tex]
[tex] \frac{-60 +/- \sqrt{ 60^{2} -4(25)(12) } }{2(25)} [/tex]
[tex] \frac{-60 +/- \sqrt{ 2400 } }{50} [/tex]
[tex] \frac{-6 +/- 2\sqrt{6} }{5} [/tex]
We can find this answer by first putting it into standard form. To do this we need to multiply the parenthesis and then solve for 0.
(5y + 6)^2 = 24
25y^2 + 60y + 36 = 24
25y^2 + 60y + 12 = 0
Now we can find the a, b and c values for the quadratic equation based on the standard form.
a = 25 (number attached to x^2)
b = 60 (number attached to x)
c = 12 (number with no variable)
Now we use these in the quadratic equation and simplify.
[tex] \frac{-b +/- \sqrt{ b^{2} -4ac } }{2a} [/tex]
[tex] \frac{-60 +/- \sqrt{ 60^{2} -4(25)(12) } }{2(25)} [/tex]
[tex] \frac{-60 +/- \sqrt{ 2400 } }{50} [/tex]
[tex] \frac{-6 +/- 2\sqrt{6} }{5} [/tex]