4.9t^2+19.8t-20=0? How do i solve this?

4.9t^2+19.8t-20=0? How do i find t ?

$$\small{\text{ Use the formula: $ \boxed{ ax^2+bx+c=0 \qquad x_{1,2}=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a} } $ }}\\\\ \small{\text{ $ t_{1,2}=\dfrac{-19.8\pm\sqrt{19.8^2-4*4.9*(-20)} }{2*4.9} $ }}\\\\ \small{\text{ $ t_{1,2}=\dfrac{-19.8\pm\sqrt{ 392.04+392 } }{9.8} $ }} }}\\\\ \small{\text{ $ t_{1,2}=\dfrac{-19.8\pm\sqrt{ 784.04 } }{9.8} $ }} }}\\\\ \small{\text{ $ t_{1,2}=\dfrac{-19.8\pm 28.001 }{9.8} $ }} }}\\\\ \small{\text{ $ t_{1}=\dfrac{-19.8 + 28.001 }{9.8} = 0.8368 $ }} }}\\\\ \small{\text{ $ t_{2}=\dfrac{-19.8 - 28.001 }{9.8} = -4.8777 $ }}$$

I'd use the quadratic formula