Solving the given equations we get,
- Solutions of quadratic equation [tex]9x^2-25=-12-10x[/tex] are x=0.77 and -1.88
- Solutions of quadratic equation [tex]4n^2-1=12n+3n^2+11[/tex] are n=12.93 and -0.93
If the quadratic equation is [tex]ax^2+bx+c=0[/tex] then the solutions of this are given by,
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Here in the problem given that,
The first quadratic equation is [tex]9x^2-25=-12-10x[/tex]
Converting it into standard form of quadratic equation we get,
[tex]9x^2-25=-12-10x\\9x^2-25+12+10x=0\\9x^2+10x-13=0[/tex]
So here a=9,b=10,c=-13
The solutions of the equation is given by,
[tex]x=\frac{-10\pm\sqrt{10^2-4\times9\times(-13)}}{2\times9}=\frac{-10\pm\sqrt{100+468}}{18}\approx\frac{-10\pm23.83}{18}[/tex]
So either,
[tex]x=\frac{-10+23.83}{18}=\frac{13.83}{18}\approx0.77[/tex]
Or,
[tex]x=\frac{-10-23.83}{18}\approx-1.88[/tex]
So the solutions are given by x=0.77 and -1.88
Second quadratic equation is [tex]4n^2-1=12n+3n^2+11[/tex]
Converting the equation in its standard form we get,
[tex]4n^2-1=12n+3n^2+11\\4n^2-1-12n-3n^2-11=0\\n^2-12n-12=0[/tex]
so here a=1,b=-12,c=-12
Solutions are given by,
[tex]n=\frac{-(-12)\pm\sqrt{(-12)^2-4\times1\times(-12)}}{2\times1}=\frac{12\pm\sqrt{144+48}}{2}\approx\frac{12\pm13.86}{2}[/tex]
So either
[tex]n=\frac{12+13.86}{2}=12.93[/tex]
Or,
[tex]n=\frac{12-13.86}{2}=-0.93[/tex]
Solutions are given by n = 12.93 and -0.93
Hence the solutions are x=0.77 and -1.88 for first equation and n=12.93 and -0.93 for second equation.
Learn more about Quadratic Equation here -
https://brainly.com/question/1214333
#SPJ10
