Answer:
Option (b) is correct.
[tex]x=\frac{5\pm\sqrt{97}}{12}[/tex]
Step-by-step explanation:
Given : [tex]5x=6x^2-3[/tex]
We have to solve using quadratic formula and find value for x.
Consider the given equation [tex]5x=6x^2-3[/tex] it can be rewritten as : [tex]6x^2-5x-3=0[/tex]
The standard form of quadratic equation is [tex]ax^2+bx+c=0[/tex] , and the quadratic formula for the standard equation is given by,
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Compare the given equation with standard equation , we have a = 6 , b = -5 and c = -3.
Substitute in quadratic formula, we have,
[tex]x=\frac{5\pm\sqrt{(5)^2-4\cdot 6\cdot (-3)}}{2(6)}[/tex]
Simplify, we get,
[tex]\Rightarrow x=\frac{5\pm\sqrt{25+72}}{12}[/tex]
[tex]\Rightarrow x=\frac{5\pm\sqrt{97}}{12}[/tex]
Thus, option (b) is correct.
Answer:
Alternative B is the correct answer.
Step-by-step explanation:
We have been given the following quadratic equation;
5x = 6x²-3. The first step is to re-arrange the equation and write the equation in standard form; ax²+bx+c = 0
6x²-5x-3 = 0 . This implies that; a = 6 , b = -5 and c = -3 . The quadratic formula is given as;
x =(-b±√b²-4ac) / 2a
Plugging in the values of a, b and c into the formula yields;
x = (-(-5)±√(-5)²-4(6)(-3) ) / 2(6)
x = (5±√25+72) / 12
x = (5±√97) / 12 which is the solution
