The solution of the problem is
Given that:
The equation is [tex]h(t)=-5t^2+14t+3[/tex] , where [tex]h(t)[/tex] is height .
The ball is [tex]8m[/tex] above the ground so [tex]h=8m[/tex] .
Now,
Substitute the value of height in given equation,
[tex]h=-5t^2+14t+3\\\\8=-5t^2+14t+3[/tex]
Subtract [tex]8[/tex] on both side to obtain the quadratic equation,
[tex]-5t^2+14t+3-8=8-8\\\\-5t^2+14t-5=0[/tex]
Multiply minus sign in both sides,
[tex]5t^2-14t+5=0[/tex]
Solve the quadratic equation ,
Where,
[tex]a=5,b=-14,c=5[/tex]
[tex]x=-b +\frac{\sqrt{b^{2}-4ac } }{2a} \\\\ x=-b -\frac{\sqrt{b^{2}-4ac } }{2a}[/tex]
Substitute the known values in the formula,
[tex]x=\frac{14+\sqrt{(-14)^2-4(5)(5)} }{2(5)} \\x=\frac{14+\sqrt{196-100} }{10} \\\\x=\frac{14+\sqrt{96} }{10} \\\\x=\frac{14+\sqrt{2*2*2*2*2*3} }{10} \\\\x=\frac{14+(4\sqrt{6}) }{10} \\\\x=\frac{7+2\sqrt{6} }{5}[/tex]
Similarly,
[tex]x=\frac{7-2\sqrt{6} }{5}[/tex]
