The formula of Sridharacharya is used. Then the solutions of the equation are 73.19 and -13.19.
It is a polynomial that is equal to zero. Polynomial of variable power 2, 1, and 0 terms are there. Any equation having one term in which the power of the variable is a maximum of 2 then it is called a quadratic equation.
The equation is given as
[tex]\rm 4\sqrt{5x + 66} = x +10[/tex]
Squaring both sides, we have
16(5x + 66) = (x + 10)²
16(5x + 66) = x² + 100 + 20x
On simplifying, we get
x² + 100 + 20x -80x - 1056 = 0
x² - 60x - 956 = 0
On solving by formula method, we have
a = 1; b = -60; and c = -965
[tex]\rm x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\\\x = \dfrac{-(-60) \pm \sqrt{(-60)^2 - 4*1*(-965)}}{2*1}\\\\\\x = 73.19, \ -13.19[/tex]
Thus, the solutions of the equation are 73.19 and -13.19.
More about the quadratic equation link is given below.
https://brainly.com/question/2263981
