If it's not what You are looking for type in the equation solver your own equation and let us solve it.
12x^2+7x+1=0
a = 12; b = 7; c = +1;
Δ = b2-4ac
Δ = 72-4·12·1
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*12}=\frac{-8}{24} =-1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*12}=\frac{-6}{24} =-1/4 $
| 3/5×f+24=4-1/5×f | | 1/x-1-2/x-2=3/2x-2-21/3/2x-4 | | 2.5x+18=4x | | 6-7x=7x-10+18 | | (2w-3)(2w+3)=0 | | 7=56÷g | | 5x+9x-92=115-9x | | 5(x+1)^2=90 | | 2x-(2x-2x-1/6)=2/3(x+2/6)-1/4 | | -2(3s-6)-12=2(7s+6)-2 | | -10x+x^2+16.75=0 | | w(w+3)=42 | | 4w^2-6w-143=0 | | 128=8*a | | |3+b|=4 | | x2+16=0 | | 0.50(x+30)+0.30x=31 | | (4x-3)(6x^2-24x+9)=0 | | (8-1)^2=x^2=(x+1)^2 | | 5(2x+5)=10x-4 | | 8x^2+22x-90=0 | | 12t=6 | | 11k-37=40 | | 4(k+2)-3K=6+9 | | 2/5(h)-7=12/5(h)-2h+3 | | 7-(3n-4)=4-4n | | 2(p+4)=32 | | 3(4x-5)^2+17=19x-8 | | 1/2x-3(x+4)=2/3 | | 59+9e=140 | | 2x-5(x-5)=-3+2x | | y=7-3× |