If it's not what You are looking for type in the equation solver your own equation and let us solve it.
h^2+19h-20=0
a = 1; b = 19; c = -20;
Δ = b2-4ac
Δ = 192-4·1·(-20)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{441}=21$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*1}=\frac{-40}{2} =-20 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*1}=\frac{2}{2} =1 $
| 36=3y-12 | | 48=3w+15 | | 5x-2+3x+70=180 | | 52=4x=12 | | (-8x+2)-(-5x+7)=0 | | q^2-24q+23=0 | | (x/4)+9=6 | | k^2+15k+14=0 | | H=-10t^2+80t+92 | | -1(1)+10=y | | s^2+48s=0 | | 4x+3(-x+10)=31 | | 7x+60=36x-56 | | x/2+12=2/3+6 | | 36+6x=9x | | -3u(u+7)=0 | | (-2x+4)-(x-11)=0 | | 3/5x+3=2x-4 | | 5v(v+9)=0 | | -x-11=2x+10= | | 2h(h+2)=0 | | 9x+6=41 | | z(3z+5)=0 | | (-2x+4)+(x-11)=0 | | 2x/3+3=-4 | | 5.5n=-60.5 | | -7/5x+3=24 | | r/3+5=14 | | -4-4x=8x+20 | | 2÷19×y+48=0 | | x/4-5/4=-12/3 | | (6r-7)/(10)=(r)/(4) |