If it's not what You are looking for type in the equation solver your own equation and let us solve it.
25x^2+50x+12=0
a = 25; b = 50; c = +12;
Δ = b2-4ac
Δ = 502-4·25·12
Δ = 1300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{13}}{2*25}=\frac{-50-10\sqrt{13}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{13}}{2*25}=\frac{-50+10\sqrt{13}}{50} $
| m2–m+46=0 | | x2–x-46=0 | | x2–x+46=0 | | 3z/10+6=-4 | | x/10-1=5 | | z/8+5=2 | | -23-17x=-9-16x | | 4t-12=4t | | 5x+14=26-3x | | -2x+15=3x+9 | | 4y+1=6y+26 | | 32=(x-7+2x+3)*2 | | 158.60=x-0.1*x | | 13y+11=8y+28 | | 4x−1=2x+13x= | | 7x+4=-2x+40x= | | 3y-12=4-y | | -10a+(-1)=0 | | 250h+20h=300 | | 10-p=4(1-p) | | 18x^2-109x+162=0 | | 3×(1,5-2,5)-(3x-5)÷0,2=12,5x | | 13-2(z-5)=(5+z)+2z-1 | | d=82 | | d=8(2) | | 2.3-1.4x=0.2 | | 2.3=1.4x=0.2 | | X+x+x+-5x=60 | | x^2+2x^2=10 | | 4x-9=3(4x+1) | | 4x-9=3(4×+1) | | 10-3/x=-9 |