Question

Solve. 10x^2 = 6 + 9x10x 2 =6+9x10, x, start superscript, 2, end superscript, equals, 6, plus, 9, x Choose 1 answer: Choose 1 answer: (Choice A) A x =\dfrac{5 \pm \sqrt{65}}{-2}x= ?2 5± 65 ? ? x, equals, start fraction, 5, plus minus, square root of, 65, end square root, divided by, minus, 2, end fraction (Choice B) B x =\dfrac{9 \pm \sqrt{321}}{20}x= 20 9± 321 ? ? x, equals, start fraction, 9, plus minus, square root of, 321, end square root, divided by, 20, end fraction (Choice C) C x =\dfrac{4 \pm \sqrt{26}}{10}x= 10 4± 26 ? ? x, equals, start fraction, 4, plus minus, square root of, 26, end square root, divided by, 10, end fraction (Choice D) D x =\dfrac{-1 \pm \sqrt{109}}{18}x= 18 ?1± 109 ? ?

Answer 1

Answer:

B

Step-by-step explanation:

I got it right

Answer 2

Answer:

Option B.

Step-by-step explanation:

If a quadratic equation is defined as

$ax^2+bx+c=0$         .... (1)

then the quadratic formula is

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

The given equation is

$10x^2=6+9x$

It can rewritten as

$10x^2-9x-6=0$            .... (2)

On comparing (1) and (2) we get

$a=10,b=-9,c=-6$

Using quadratic formula we get

$x=\dfrac{-(-9)\pm \sqrt{(-9)^2-4(10)(-6)}}{2(10)}$

$x=\dfrac{9\pm \sqrt{81+240}}{20}$

$x=\dfrac{9\pm \sqrt{321}}{20}$

Therefore, the correct option is B.