If it's not what You are looking for type in the equation solver your own equation and let us solve it.
t^2+t-1.2=0
a = 1; b = 1; c = -1.2;
Δ = b2-4ac
Δ = 12-4·1·(-1.2)
Δ = 5.8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{5.8}}{2*1}=\frac{-1-\sqrt{5.8}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{5.8}}{2*1}=\frac{-1+\sqrt{5.8}}{2} $
| f-13=22 | | 2(4s-1)=-2(-8+2s) | | 2x^2-1=2^x | | 9x=+210=4x+140 | | 2(2x+4)+10=x | | 3(x-2)+7=5-(x+3) | | 3r-7=r+8 | | 2(3x-6)=-3(2x+4) | | x2=9/625 | | 2/3x+8=1/3-2 | | 1-w=6-6w | | 20-c=10c-2 | | 6L+20s=270(-2) | | 6L+20s=270 | | 4(9-x)=3(2x+2) | | 2(a+)-7=9 | | .45x=36 | | 8^1+1=16(2^a-1) | | f-13=22/35 | | 49x3-x=0 | | 4(3x-2)=33 | | f-13=22;35 | | 1290=h/10+h/5 | | a+8=5;0 | | 5(6x-4)=140 | | m=4=2m | | x²+15x=5 | | 3x+2/3=2x+3/4 | | (5x+6)^2=9 | | --a=7a | | X+(1/2x)=120 | | 3a+a-7a=25 |