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Hi my lil bunny!
❧⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯☙
Lets do this step by step.
[tex]30x^{2\ }-\ 28x\ +\ 6\ =\ 0[/tex]
Use the quadratic formula to find the solutions.
[tex]\frac{-b± \sqrt{ b^2 - 4 (ac)} }{2a}[/tex]
Substitute the values [tex]a = 30, b = -28 ,[/tex] and [tex]c = 6[/tex] into the quadratic formula and solve for [tex]x.[/tex]
[tex]\frac{28±\sqrt{( -28)^2 -4 . (30 . 6)} }{2 . 30}[/tex]
Simplify.
__________
Simplify the numerator.
____________
Raise [tex]-28[/tex] to the power of 2.
[tex]\frac{28±\sqrt{ 784 -4 . (30 . 6)} }{2 . 30}[/tex]
Multiply 30 by 6.
[tex]x = \frac{28±\sqrt{ 784 - 4 . 180} }{2 . 30}[/tex]
Multiply -4 by 180.
[tex]x = \frac{28±\sqrt{ 784 - 720} }{2 . 30}[/tex]
Subtract 720 from 784.
[tex]x = \frac{28 ±\sqrt{8^2} }{2 . 30}[/tex]
Rewrite 64 as 8^2.
[tex]x = \frac{28 ± 8}{2 . 30}[/tex]
Pull terms out from under the radical, assuming positive real numbers.
[tex]x = \frac{28 ± 8}{2 . 30}[/tex]
Multiply 2 by 30.
[tex]x = \frac{28 ± 8}{60}[/tex]
Simplify [tex]\frac{28 ± 8}{60}[/tex]
[tex]x = \frac{7 ± 2}{15}[/tex]
The final answer is the combination of both solutions.
[tex]x = \frac{1}{3} , \frac{3}{5}[/tex]
❧⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯☙
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Hope this helped you.
Could you maybe give brainliest..?
❀*May*❀
