If it's not what You are looking for type in the equation solver your own equation and let us solve it.
7t^2-2=14t
We move all terms to the left:
7t^2-2-(14t)=0
a = 7; b = -14; c = -2;
Δ = b2-4ac
Δ = -142-4·7·(-2)
Δ = 252
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6\sqrt{7}}{2*7}=\frac{14-6\sqrt{7}}{14} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6\sqrt{7}}{2*7}=\frac{14+6\sqrt{7}}{14} $
| 115-m=180 | | 3m^2-12m+10=0 | | 2*3.14*28.7=c | | -x+188=275 | | 115-x=190 | | 30=20+50*10^-0.04m | | 7(x+5)-2x-15=2x-10+3(2x+10)-2(x-5) | | 5/4(2-k)=2(3k-1)-2/4k | | (-4z-1)-(-7z+5)=6z-12 | | 4/5d-2=1/5d+1 | | 6n+2(-5n+6)=40 | | 3x+7x+22=180 | | 3(-7t+5)=-(-6t+336) | | 2x+10+4x+12=180 | | 1/5z+3=-z+21 | | 50+4x+12+2x+10=180 | | -4r+1=-15 | | 1/5n+2=-70 | | -46=-6-5c | | 4/3d-1=-9 | | 50+2x+10+4x+12=180 | | 4/3d-1=0 | | -1/8x-2=0 | | r-40=-20 | | 5(x-3=-4x+30 | | 4x+40=-8(x+7) | | 5(x-1)=-7x+43 | | x=75-0.3x | | -4c+3=23 | | 7v+8(v+2)=-29 | | 7(w-2)-3w=-30 | | 9=-6x+3(x+7) |