Answer:
[tex]\boxed{\textsf{ The value of x is }\bf{ \dfrac{-5\pm\sqrt{21}}{2}}.}[/tex]
Step-by-step explanation:
A quadratic equation is given to us and we need to find the Solutions of the quadratic equation . So the given equⁿ is ,
[tex]\sf\implies (x+4)^2 -3(x+4) -3 = 0 [/tex]
Firstly let's simpify the quadratic equation and then use the quadratic formula also known as " Shreedhacharya's Formula" .
Taking the given equation :-
[tex]\sf\implies (x+4)^2 -3(x+4) -3 = 0\\\\\sf\implies x^2 + 16 + 8x -3x -12 -3 = 0 \\\\\sf\implies x^2+5x +1 = 0 [/tex]
Now let's use the Quadratic formula ,
Quadratic Formula :-
[tex]\qquad\boxed{\boxed{ \sf x =\dfrac{ -b \pm \sqrt{ b^2-4ac}}{2a} }} [/tex]
With respect to Standard form ax² + bx + c form. Here a = 1 , b = 5 and c = 1 .
[tex]\sf\implies x = \dfrac{ -5\pm \sqrt{ (5)^2-4(1)(1)}}{2(1)} \\\\\sf\implies x = \dfrac{ -5 \pm \sqrt{ 25 -4 } }{2} \\\\\sf\implies \boxed{\pink{\sf x = \dfrac{-5+\sqrt{21}}{2}, \dfrac{-5-\sqrt{21}}{2} }}[/tex]
