The solutions of the given equation are x= (√5+3)/2 & x= (-√5+3)/2
If ax²+bx+c = 0 is a quadratic equation ,where a,b,c are real numbers, Then the solution is x = (-b±√(b²-4ac))/2a , where a≠0
The given equation is, 4x²-12x+9 = 5
So, 4x²-12x+9 = 5
⇒ 4x²-12x+9-5 = 0
⇒ 4x²-12x+4 = 0
Comparing this equation with ax²+bx+c = 0 we get,
a = 4, b = -12, c = 4
Applying sridharacharya formula we get,
x = (-(-12)±√((-12)²-4×4×4))/(2×4)
= (12±√(144-64)/8
= (12±√80)/8
= (3±√5)/2
The values of x are (√5+3)/2 & (-√5+3)/2
Learn more about sridharacharya formula here :
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